ADVANCED INTEGRATION QUESTIONS AND SOLUTIONS
Evaluate the integral ∫3x2 dx\int 3x^2 \, dx∫3x2dx
To integrate 3x23x^23x2, we use the power rule of integration, which states that ∫xn
dx=xn+1n+1+C\int x^n \, dx = \frac{x^{n+1}}{n+1} + C∫xndx=n+1xn+1+C for any
constant n≠−1n \neq -1n =−1.
∫3x2 dx=3∫x2 dx=3(x33)+C=x3+C\int 3x^2 \, dx = 3 \int x^2 \, dx = 3 \left( \frac{x^3}
{3} \right) + C = x^3 + C∫3x2dx=3∫x2dx=3(3x3)+C=x3+C
2. Find the integral ∫(4x3−2x+1) dx\int (4x^3 - 2x + 1) \, dx∫(4x3−2x+1)dx
We integrate each term separately using the power rule of integration.
∫(4x3−2x+1) dx=∫4x3 dx−∫2x dx+∫1 dx\int (4x^3 - 2x + 1) \, dx = \int 4x^3 \, dx - \int
2x \, dx + \int 1 \, dx∫(4x3−2x+1)dx=∫4x3dx−∫2xdx+∫1dx
Integrate each term:
∫4x3 dx=4(x44)=x4\int 4x^3 \, dx = 4 \left( \frac{x^4}{4} \right) = x^4∫4x3dx=4(4x4
)=x4 ∫2x dx=2(x22)=x2\int 2x \, dx = 2 \left( \frac{x^2}{2} \right) = x^2∫2xdx=2(2x2
)=x2 ∫1 dx=x\int 1 \, dx = x∫1dx=x
Combine the results:
∫(4x3−2x+1) dx=x4−x2+x+C\int (4x^3 - 2x + 1) \, dx = x^4 - x^2 + x +
C∫(4x3−2x+1)dx=x4−x2+x+C
3. Determine the integral ∫ex dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\
I = -e^{2x} \cos(x) + 2 \left( e^{2x} \sin(x) - 2I \right) \\
\text{Simplify and solve for } I: \\
I = -e^{2x} \cos(x) + 2e^{2x} \sin(x) - 4I \\
\text{Combine like terms:} \\
5I = -e^{2x} \cos(x) + 2e^{2x} \sin(x) \\
I = \frac{1}{5} \left( -e^{2x} \cos(x) + 2e^{2x} \sin(x) \right) \\
\text{So:} \\
\int e^{2x} \sin(x) \, dx = \frac{1}{5} \left( -e^{2x} \cos(x) + 2e^{2x} \sin(x) \right) \\
\text{Now, use this result to integrate } \int x^2 e^{2x} \sin(x) \, dx: \\
\text{Set:} \\
u = x^2, \quad dv = \left( \frac{1}{5} \left( -e^{2x} \cos(x) + 2e^{2x} \sin(x) \right) \
right) dx \\
\text{Then:} \\
3. Determine the integral ∫ex dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
3. Determine the integral ∫ex dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\v3. Determine the integral
∫ex dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
3. Determine the integral ∫ex dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\\int e^{2x} \sin(x) \, dx = -
e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\3. Determine the integral ∫ex
dx\int e^x \, dx∫exdx
The integral of the exponential function exe^xex is straightforward because it is its
own derivative.
∫ex dx=ex+C\int e^x \, dx = e^x + C∫exdx=ex+C
4. Calculate the integral ∫sin (x) dx\int \sin(x) \, dx∫sin(x)dx
The integral of sin (x)\sin(x)sin(x) is −cos (x)-\cos(x)−cos(x), since the derivative of
−cos (x)-\cos(x)−cos(x) is sin (x)\sin(x)sin(x).
∫sin (x) dx=−cos (x)+C\int \sin(x) \, dx = -\cos(x) + C∫sin(x)dx=−cos(x)+C
5. Evaluate the integral ∫1x dx\int \frac{1}{x} \, dx∫x1dx
The integral of 1x\frac{1}{x}x1 is the natural logarithm of the absolute value of xxx.
∫1x dx=ln ∣x∣+C\int \frac{1}{x} \, dx = \ln |x| + C∫x1dx=ln∣x∣+C
\begin{aligned}
\text{Evaluate the integral } \int x^2 e^{2x} \sin(x) \, dx. \\
\text{Solution} \\
\text{To solve the integral } \int x^2 e^{2x} \sin(x) \, dx, \text{ we will use integration
by parts multiple times.} \\
\text{First, we recall the integration by parts formula:} \\
\int u \, dv = uv - \int v \, du \\
\text{Let's set } u = x^2 \text{ and } dv = e^{2x} \sin(x) \, dx. \\
\text{Then we have:} \\
du = 2x \, dx \\
\text{We need to find } v \text{ by integrating } dv. \text{ To do this, we need to
evaluate the integral } \int e^{2x} \sin(x) \, dx. \text{ This integral requires integration
by parts itself.} \\
\text{Let's solve } \int e^{2x} \sin(x) \, dx \text{ using integration by parts again.} \\
\text{Set:} \\
u_1 = e^{2x}, \quad dv_1 = \sin(x) \, dx \\
\text{Then:} \\
du_1 = 2e^{2x} \, dx, \quad v_1 = -\cos(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \sin(x) \, dx: \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) - \int (-\cos(x)) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \sin(x) \, dx = -e^{2x} \cos(x) + 2 \int e^{2x} \cos(x) \, dx \\
\text{Now, we need to integrate } \int e^{2x} \cos(x) \, dx. \text{ Using integration by
parts again:} \\
\text{Set:} \\
u_2 = e^{2x}, \quad dv_2 = \cos(x) \, dx \\
\text{Then:} \\
du_2 = 2e^{2x} \, dx, \quad v_2 = \sin(x) \\
\text{Using the integration by parts formula for } \int e^{2x} \cos(x) \, dx: \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - \int \sin(x) \cdot 2e^{2x} \, dx \\
\text{Simplify the integral:} \\
\int e^{2x} \cos(x) \, dx = e^{2x} \sin(x) - 2 \int e^{2x} \sin(x) \, dx \\
\text{Let } \int e^{2x} \sin(x) \, dx = I. \text{ Then:} \\
du = 2x \, dx, \quad v = \frac{1}{5} \left( -e^{2x} \cos(x) + 2e^{2x} \sin(x) \right) \\
\text{Using integration by parts formula:} \\
\int x^2 e^{2x} \sin(x) \, dx = x^2 \cdot \frac{1}{5} \left( -e^{2x} \cos(x) + 2e^{2x} \
sin(x) \right) - \int \left( \frac{1}{5} \left( -e^{2x} \cos(x) + 2e^{2x} \sin(x) \right) \
cdot 2x \right) dx \\
\text{Simplify:} \\
= \frac{1}{5} x^2 \left( -e^{2x} \cos(x) + 2e^{2x} \sin(x) \right) - \frac{2}{5} \int x \
left( -e^{2x} \cos(x) + 2e^{2x} \sin(x) \right) dx \\
\text{This integral can be further simplified and solved, but it is quite complex. To
solve it fully, we would continue using integration by parts and substitution
techniques until all terms are integrated.} \\
\text{The integral } \int x^2 e^{2x} \sin(x) \, dx \text{ is indeed quite intricate and
illustrates the need for multiple techniques in integration.}
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