MATH 132 - CALCULUS AND
ANALYTIC GEOMETRY II -
Integration by Parts
Question Bank - Set 2
Liberty University
Question 1
Question
Evaluate the integral ∫xexdx using integration by parts.
Solution
To evaluate the integral ∫xexdx using integration by parts, we will use the
formula: ∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let’s choose u=xand dv =exdx. Then, we have:
du =dx and v=∫exdx =ex
Step 2: Now, we can apply the integration by parts formula:
∫xexdx =uv −∫v du
=x·ex−∫exdx
=x·ex−ex+C
Therefore, the integral ∫xexdx evaluates to xex−ex+C, where Cis the
constant of integration.
Question 2
Question
Evaluate the integral ∫xsin−1(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xsin−1(x)dx using integration by parts, we will
let u= sin−1(x)and dv =x dx. Then, we have du =1
√1−x2dx and v=1
2x2.
Step 1: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 2: Substitute u,dv,du, and vinto the formula:
∫xsin−1(x)dx =1
2x2sin−1(x)−∫1
2x2·1
√1−x2dx
Step 3: Compute the new integral:
∫1
2x2·1
√1−x2dx =1
2∫x2
√1−x2dx
Step 4: Let x= sin(t), then dx = cos(t)dt
1
2∫x2
√1−x2dx =1
2∫sin2(t)
cos(t)cos(t)dt =1
2∫sin2(t)dt
Step 5: Use the trigonometric identity sin2(t) = 1
2−1
2cos(2t)to simplify
the integral: 1
2∫sin2(t)dt =1
4t−1
4sin(2t) + C
Step 6: Substitute back xfor tand combine the results:
∫xsin−1(x)dx =1
2x2sin−1(x)−1
4arcsin(x) + 1
4x√1−x2+C
Therefore, ∫xsin−1(x)dx =1
2x2sin−1(x)−1
4arcsin(x) + 1
4x√1−x2+C
where Cis the constant of integration.
Question 3
Question
Evaluate the integral ∫excos x dx using integration by parts.
2
Solution
To evaluate ∫excos x dx using integration by parts, we will apply the formula
∫u dv =uv −∫v du. Let u=exand dv = cos x dx. Then, we have du =exdx
and v=∫cos x dx = sin x.
Step 1: Apply the integration by parts formula:
∫excos x dx =exsin x−∫sin x·exdx
=exsin x−(−∫exsin x dx)
Step 2: Apply integration by parts again to evaluate ∫exsin x dx: Let
u=exand dv = sin x dx. Then, du =exdx and v=−cos x.
∫exsin x dx =−excos x−∫(−cos x)·exdx
=−excos x−(−∫excos x dx)
Step 3: Substitute ∫excos x dx back into the equation from Step 2:
∫exsin x dx =−excos x−(−exsin x+∫excos x dx)
∫exsin x dx =−excos x+exsin x−∫excos x dx
Step 4: Substitute the result back into the original integral:
∫excos x dx =exsin x−(−excos x+exsin x−∫excos x dx)
2∫excos x dx = 2exsin x−excos x
∫excos x dx =2exsin x−excos x
2
Therefore, the solution is ∫excos x dx =2exsin x−excos x
2+C, where Cis the
constant of integration.
Question 4
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
3
Solution
To evaluate the integral ∫x2ln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x2dx. Then, we will differentiate uto find du and integrate
dv to find v.
Step 1: Let u= ln(x)and dv =x2dx. Then, we have:
du =1
xdx and v=1
3x3
Step 2: Apply integration by parts formula:
∫x2ln(x)dx =uv −∫v du
= ln(x)·1
3x3−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
Step 3: Simplify the integral:
∫x2ln(x)dx =1
3x3ln(x)−1
3∫x2dx
=1
3x3ln(x)−1
3·1
3x3+C
=1
3x3ln(x)−1
9x3+C
Therefore, ∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C, where Cis the constant of
integration.
Question 5
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will let u=
ln(x)and dv =x dx.
Step 1: Compute du and v.
• Since u= ln(x), differentiate uwith respect to xto find du:
du
dx =1
x
du =1
xdx
4
• Since dv =x dx, integrate dv with respect to xto find v:
v=∫x dx =x2
2
Step 2: Apply the integration by parts formula: ∫u dv =uv −∫v du.
∫xln(x)dx = ln(x)·x2
2−∫x2
2·1
xdx
=x2ln(x)
2−1
2∫x dx
=x2ln(x)
2−x2
4+C
where Cis the constant of integration.
Therefore, ∫xln(x)dx =x2ln(x)
2−x2
4+C.
Question 6
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will use the
formula: ∫u dv =uv −∫v du.
Step 1: Choose uand dv. Let u= ln(x)and dv =x dx.
Step 2: Calculate du and v. Calculate du by taking the derivative of uwith
respect to x:
du =1
xdx
Calculate vby integrating dv with respect to x:
v=1
2x2
Step 3: Apply the integration by parts formula. Using the formula ∫u dv =
uv −∫v du, we have:
∫xln(x)dx = (ln(x))( 1
2x2)−∫1
2x2·1
xdx
Step 4: Simplify and evaluate the integral. Simplify the expression:
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
5
Integrating the last term:
∫x dx =1
2x2
Step 5: Combine the terms. Putting it all together, we have:
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
where Cis the constant of integration.
Question 7
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xln(x)dx using integration by parts, we will
apply the formula:
∫u dv =uv −∫v du
where we choose u= ln(x)and dv =x dx.
Step 1: Find du and v.
du =1
xdx
v=1
2x2
Step 2: Apply integration by parts formula.
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
Step 3: Evaluate the integral.
∫xln(x)dx =1
2x2ln(x)−1
2·1
2x2+C
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
6
Question 8
Question
Evaluate the integral ∫exsin x dx using integration by parts.
Solution
To evaluate ∫exsin x dx using integration by parts, we will choose u=exand
dv = sin x dx.
Step 1: Calculate du and v.
Let u=ex(Choose u=ex)
⇒du =exdx (Calculate du)
Let dv = sin x dx (Choose dv = sin x dx)
⇒v=−cos x(Integrate dv)
Step 2: Apply the integration by parts formula ∫u dv =uv −∫v du.
∫exsin x dx =ex(−cos x)−∫(−cos x)(exdx)
=−excos x+∫excos x dx
Step 3: Apply integration by parts again to evaluate ∫excos x dx. Let
u=exand dv = cos x dx.
Step 4: Calculate du and v.
Let u=ex(Choose u=ex)
⇒du =exdx (Calculate du)
Let dv = cos x dx (Choose dv = cos x dx)
⇒v= sin x(Integrate dv)
Step 5: Apply the integration by parts formula again.
∫excos x dx =ex(sin x)−∫(sin x)(exdx)
=exsin x−∫exsin x dx
Step 6: Substitute the result back into the original integral.
∫exsin x dx =−excos x+exsin x−∫exsin x dx
7
Step 7: Solve for ∫exsin x dx.
2∫exsin x dx =−excos x+exsin x
∫exsin x dx =1
2(exsin x−excos x) + C
Therefore, ∫exsin x dx =1
2(exsin x−excos x) + C.
Question 9
Question
Evaluate the definite integral ∫π
4
0xcos(x)dx using integration by parts.
Solution
To solve the given integral, we will use integration by parts. Integration by
parts formula is given by:
∫u dv =uv −∫v du
where uand vare differentiable functions of x. In this case, we will let u=x
and dv = cos(x)dx. Then, we will differentiate uto find du and integrate dv to
find v.
Step 1: Find du and v
u=x, dv = cos(x)dx
Taking the differential of ugives:
du =dx
Integrating dv gives:
v=∫cos(x)dx = sin(x)
Step 2: Apply the integration by parts formula Using the formula ∫u dv =
uv −∫v du, we have:
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C
8
Step 3: Evaluate the definite integral Now, we can evaluate the definite
integral ∫π
4
0xcos(x)dx using the antiderivative we found:
∫π
4
0
xcos(x)dx = [xsin(x) + cos(x)]
π
4
0
=(π
4sin (π
4)+ cos (π
4))−(0 sin(0) + cos(0))
=(π
4·
√2
2+√2
2)−(0 + 1)
=π√2
8+√2
2−1
Therefore, ∫π
4
0xcos(x)dx =π√2
8+√2
2−1.
Question 10
Question
Evaluate the integral ∫x2ln x dx using integration by parts.
Solution
To solve this integral, we will use integration by parts, which states:
∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let’s choose u= ln xand dv =x2dx. Then, we have:
du =1
xdx
v=∫x2dx =x3
3
Step 2: Now, we can apply the integration by parts formula:
∫x2ln x dx =x3
3ln x−∫x3
3·1
xdx
Simplify:
∫x2ln x dx =x3
3ln x−1
3∫x2dx
Step 3: Integrate the remaining term:
∫x2ln x dx =x3
3ln x−1
3·x3
3+C
9
∫x2ln x dx =x3
3ln x−x3
9+C
Therefore, the solution to the integral ∫x2ln x dx is x3
3ln x−x3
9+C, where
Cis the constant of integration.
Question 11
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To integrate the given function ∫x2exdx, we will use integration by parts, which
states: ∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Choose uand dv. Let u=x2and dv =exdx.
Step 2: Calculate du and v.
du =d
dx (x2) = 2x dx
To find v, we need to integrate dv:
v=∫exdx =ex
Step 3: Apply the integration by parts formula.
∫x2exdx =x2ex−∫2xexdx
Step 4: Integrate the remaining integral. Using integration by parts again
with u=xand dv =exdx:
du =d
dx (x) = dx
v=∫exdx =ex
Substitute these values into the formula:
∫2xexdx = 2xex−∫2exdx
Step 5: Evaluate the final integral.
∫2exdx = 2ex
10
Step 6: Combine all the terms. Putting everything together, we have:
∫x2exdx =x2ex−2xex+ 2ex+C
where Cis the constant of integration.
Question 12
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To evaluate the integral ∫x2exdx using integration by parts, we will let u=x2
and dv =exdx. Then, we will differentiate uto find du and integrate dv to find
v. Finally, we will apply the integration by parts formula ∫u dv =uv −∫v du.
Step 1: Let u=x2and dv =exdx.Step 2: Differentiate uto find du:
du = 2x dx
Step 3: Integrate dv to find v:
v=∫exdx =ex
Step 4: Apply the integration by parts formula:
∫x2exdx =x2ex−∫ex·2x dx
∫x2exdx =x2ex−2∫xexdx
Step 5: We now have a new integral ∫xexdx. Let’s use integration by parts
again on this integral.
Step 6: Let u=xand dv =exdx.Step 7: Differentiate uto find du:
du =dx
Step 8: Integrate dv to find v:
v=ex
Step 9: Apply the integration by parts formula again:
∫xexdx =xex−∫exdx
∫xexdx =xex−ex
11
Step 10: Substitute back into the previous integral:
∫x2exdx =x2ex−2(xex−ex) + C
∫x2exdx =x2ex−2xex+ 2ex+C
Answer: ∫x2exdx =x2ex−2xex+ 2ex+C
Question 13
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will let u=
ln(x)and dv =x dx.
Step 1: Compute du and v.
du =1
xdx and v=1
2x2
Step 2: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 3: Substitute u,dv,v, and du into the formula:
= ln(x)·1
2x2−∫1
2x2·1
xdx
Step 4: Simplify the integral:
=1
2x2ln(x)−1
2∫x dx
Step 5: Integrate the remaining integral:
=1
2x2ln(x)−1
2·1
2x2+C
Step 6: Simplify the final result:
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
12
Question 14
Question
Evaluate the integral ∫xln x dx using integration by parts.
Solution
To evaluate the integral ∫xln x dx using integration by parts, we will choose
u= ln xand dv =x dx. We will then differentiate uto get du and integrate dv
to get v.
u= ln x
dv =x dx
du =1
xdx
v=1
2x2
Step 1: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 2: Substitute u,dv,v, and du into the formula:
∫xln x dx =1
2x2ln x−∫1
2x2·1
xdx
=1
2x2ln x−1
2∫x dx
Step 3: Evaluate the remaining integral:
∫x dx =1
2x2+C
Step 4: Substitute back the value of the integral and simplify:
∫xln x dx =1
2x2ln x−1
2(1
2x2)+C
=1
2x2ln x−1
4x2+C
Therefore, ∫xln x dx =1
2x2ln x−1
4x2+C, where Cis the constant of
integration.
13
Question 15
Question
Evaluate the definite integral:
∫1
0
xln(x)dx
Solution
To evaluate the given integral, we will use integration by parts by choosing
u= ln(x)and dv =x dx. Then, we will use the formula:
∫u dv =uv −∫v du
Step 1: Determine du and v.
•du =1
xdx
•v=x2
2
Step 2: Apply integration by parts.
∫1
0
xln(x)dx =[x2
2ln(x)]1
0−∫1
0
x2
2·1
xdx
=[1
2ln(1) −0]−1
2∫1
0
x dx
= 0 −1
4=−1
4
Therefore, ∫1
0xln(x)dx =−1
4.
Question 16
Question
Evaluate the integral ∫excos(x)dx using integration by parts.
Solution
To evaluate the integral ∫excos(x)dx using integration by parts, we will choose
u=exand dv = cos(x)dx. Then, we will differentiate uto get du and integrate
dv to get v.
14
Step 1: Let u=exand dv = cos(x)dx. Then, du =exdx and v=
∫cos(x)dx = sin(x).
Step 2: Apply the integration by parts formula: ∫u dv =uv −∫v du.
∫excos(x)dx =exsin(x)−∫sin(x)·exdx
=exsin(x)−(−excos(x)−∫−excos(x)dx)
=exsin(x) + excos(x)−∫excos(x)dx
Step 3: Rearrange the equation to solve for ∫excos(x)dx.
2∫excos(x)dx =exsin(x) + excos(x)
∫excos(x)dx =1
2(exsin(x) + excos(x))
Therefore, the integral ∫excos(x)dx is 1
2(exsin(x) + excos(x)) + C, where
Cis the constant of integration.
Question 17
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x dx. Then, we will differentiate uto find du and integrate
dv to find v.
Step 1: Find du and v.
•u= ln(x)
•du =1
xdx
•dv =x dx ⇒v=1
2x2
Step 2: Apply the formula for integration by parts:
∫u dv =uv −∫v du
15
Step 3: Substitute u,v,du, and dv into the integration by parts formula:
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
Step 4: Simplify the right-hand side of the equation:
1
2x2ln(x)−∫1
2x dx
Step 5: Integrate the remaining integral:
1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 18
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we use the for-
mula ∫u dv =uv −∫v du.
Step 1: Identify uand dv Let u= ln(x)and dv =x dx. Then, calculate
du and v.
•du =1
xdx
• To find v, we integrate dv:
v=∫x dx =1
2x2
Step 2: Apply the formula Now, we apply the integration by parts
formula: ∫xln(x)dx =uv −∫v du
∫xln(x)dx = ln(x)·1
2x2−∫1
2x2·1
xdx
Step 3: Simplify and integrate
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
Thus, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of inte-
gration.
16
Question 19
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xln(x)dx using integration by parts, we will
choose u= ln(x)and dv =x dx, so that du =1
xdx and v=1
2x2. Now we will
apply the integration by parts formula:
∫u dv =uv −∫v du
Step 1: Let’s calculate du and v:
•du =1
xdx
•v=1
2x2
Step 2: Now we can apply the integration by parts formula:
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
2·1
2x2+C
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 20
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx, we will use integration by parts. Recall
the formula for integration by parts:
∫u dv =uv −∫v du
17
Step 1: Let’s choose uand dv. Let u= ln(x)and dv =x dx. Then, we
have:
du =1
xdx
v=1
2x2
Step 2: Now, let’s apply the integration by parts formula:
∫xln(x)dx =uv −∫v du
= ln(x)·1
2x2−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 21
Question
Evaluate the integral ∫xsin(3x)dx using integration by parts.
Solution
To evaluate the integral ∫xsin(3x)dx, we will use integration by parts. The
formula for integration by parts is given by:
∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Identify uand dv
Let u=xand dv = sin(3x)dx. Then, calculate du and v:
du =dx
v=−1
3cos(3x)
Step 2: Apply integration by parts formula
Using the integration by parts formula, we have:
∫xsin(3x)dx =uv −∫v du
18
Step 3: Calculate the integral
=x(−1
3cos(3x))−∫(−1
3cos(3x))dx
=−x
3cos(3x) + 1
3∫cos(3x)dx
Step 4: Evaluate the remaining integral
=−x
3cos(3x) + 1
3(1
3sin(3x))+C
Step 5: Final answer
Therefore, the solution to the integral ∫xsin(3x)dx is:
−x
3cos(3x) + 1
9sin(3x) + C
where Cis the constant of integration.
Question 22
Question
Evaluate the integral ∫xsin(2x)dx using integration by parts.
Solution
To evaluate the integral ∫xsin(2x)dx using integration by parts, we will set:
u=xand dv = sin(2x)dx
Taking the differentials, we get:
du =dx and v=−1
2cos(2x)
Step 1: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 2: Substitute u,dv,du, and vinto the integration by parts formula:
∫xsin(2x)dx =x(−1
2cos(2x))−∫(−1
2cos(2x))dx
=−1
2xcos(2x) + 1
2∫cos(2x)dx
19
Step 3: Integrate ∫cos(2x)dx with respect to x:
∫cos(2x)dx =1
2sin(2x) + C
Step 4: Substitute the result back into the equation:
−1
2xcos(2x) + 1
2(1
2sin(2x))+C
Therefore, the solution to the integral ∫xsin(2x)dx is −1
2xcos(2x)+1
4sin(2x)+
C, where Cis the constant of integration.
Question 23
Question
Evaluate the integral ∫exsin x dx using integration by parts.
Solution
To evaluate the integral ∫exsin x dx using integration by parts, we will choose
u= sin xand dv =exdx.
Step 1: Compute du and v.
•du =d
dx (sin x)dx = cos x dx
•v=∫exdx =ex
Step 2: Apply the integration by parts formula, ∫u dv =uv −∫v du.
∫exsin x dx = sin x·ex−∫excos x dx
Step 3: The remaining integral ∫excos x dx can be evaluated using inte-
gration by parts again. Let’s choose u= cos xand dv =exdx.
Step 4: Compute du and v.
•du =d
dx (cos x)dx =−sin x dx
•v=∫exdx =ex
Step 5: Apply the integration by parts formula once more.
∫excos x dx = cos x·ex−∫ex(−sin x)dx
= cos x·ex+∫exsin x dx
20
Step 6: Substitute the result from Step 5 back into the initial integration
by parts formula.
∫exsin x dx = sin x·ex−(cos x·ex+∫exsin x dx)
2∫exsin x dx = (sin x−cos x)·ex
∫exsin x dx =(sin x−cos x)·ex
2+C
Therefore, ∫exsin x dx =(sin x−cos x)·ex
2+C.
Question 24
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the given integral, we will use the formula for integration by parts:
∫u dv =uv −∫v du
Let’s choose u= ln(x)and dv =x2dx. Then, we have du =1
xdx and v=1
3x3.
Step 1: Apply integration by parts:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
Step 2: Evaluate the remaining integral:
∫x2dx =1
3x3+C
Step 3: Substitute back into the original integral:
∫x2ln(x)dx =1
3x3ln(x)−1
3(1
3x3)+C
=1
3x3ln(x)−1
9x3+C
Therefore, the integral ∫x2ln(x)dx evaluates to 1
3x3ln(x)−1
9x3+C, where
Cis the constant of integration.
21
Question 25
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we let u= ln(x)
and dv =x dx. Then, du =1
xdx and v=1
2x2.
Step 1: Apply integration by parts formula: ∫u dv =uv −∫v du.
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
Step 2: Integrate the remaining term.
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C
So, the integral ∫xln(x)dx evaluates to 1
2x2ln(x)−1
4x2+C, where Cis the
constant of integration.
22
Question 2
Question
Evaluate the integral ∫xsin−1(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xsin−1(x)dx using integration by parts, we will
let u= sin−1(x)and dv =x dx. Then, we have du =1
√1−x2dx and v=1
2x2.
Step 1: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 2: Substitute u,dv,du, and vinto the formula:
∫xsin−1(x)dx =1
2x2sin−1(x)−∫1
2x2·1
√1−x2dx
Step 3: Compute the new integral:
∫1
2x2·1
√1−x2dx =1
2∫x2
√1−x2dx
Step 4: Let x= sin(t), then dx = cos(t)dt
1
2∫x2
√1−x2dx =1
2∫sin2(t)
cos(t)cos(t)dt =1
2∫sin2(t)dt
Step 5: Use the trigonometric identity sin2(t) = 1
2−1
2cos(2t)to simplify
the integral: 1
2∫sin2(t)dt =1
4t−1
4sin(2t) + C
Step 6: Substitute back xfor tand combine the results:
∫xsin−1(x)dx =1
2x2sin−1(x)−1
4arcsin(x) + 1
4x√1−x2+C
Therefore, ∫xsin−1(x)dx =1
2x2sin−1(x)−1
4arcsin(x) + 1
4x√1−x2+C
where Cis the constant of integration.
Question 3
Question
Evaluate the integral ∫excos x dx using integration by parts.
2
Solution
To evaluate ∫excos x dx using integration by parts, we will apply the formula
∫u dv =uv −∫v du. Let u=exand dv = cos x dx. Then, we have du =exdx
and v=∫cos x dx = sin x.
Step 1: Apply the integration by parts formula:
∫excos x dx =exsin x−∫sin x·exdx
=exsin x−(−∫exsin x dx)
Step 2: Apply integration by parts again to evaluate ∫exsin x dx: Let
u=exand dv = sin x dx. Then, du =exdx and v=−cos x.
∫exsin x dx =−excos x−∫(−cos x)·exdx
=−excos x−(−∫excos x dx)
Step 3: Substitute ∫excos x dx back into the equation from Step 2:
∫exsin x dx =−excos x−(−exsin x+∫excos x dx)
∫exsin x dx =−excos x+exsin x−∫excos x dx
Step 4: Substitute the result back into the original integral:
∫excos x dx =exsin x−(−excos x+exsin x−∫excos x dx)
2∫excos x dx = 2exsin x−excos x
∫excos x dx =2exsin x−excos x
2
Therefore, the solution is ∫excos x dx =2exsin x−excos x
2+C, where Cis the
constant of integration.
Question 4
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
3
Solution
To evaluate the integral ∫x2ln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x2dx. Then, we will differentiate uto find du and integrate
dv to find v.
Step 1: Let u= ln(x)and dv =x2dx. Then, we have:
du =1
xdx and v=1
3x3
Step 2: Apply integration by parts formula:
∫x2ln(x)dx =uv −∫v du
= ln(x)·1
3x3−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
Step 3: Simplify the integral:
∫x2ln(x)dx =1
3x3ln(x)−1
3∫x2dx
=1
3x3ln(x)−1
3·1
3x3+C
=1
3x3ln(x)−1
9x3+C
Therefore, ∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C, where Cis the constant of
integration.
Question 5
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will let u=
ln(x)and dv =x dx.
Step 1: Compute du and v.
• Since u= ln(x), differentiate uwith respect to xto find du:
du
dx =1
x
du =1
xdx
4
• Since dv =x dx, integrate dv with respect to xto find v:
v=∫x dx =x2
2
Step 2: Apply the integration by parts formula: ∫u dv =uv −∫v du.
∫xln(x)dx = ln(x)·x2
2−∫x2
2·1
xdx
=x2ln(x)
2−1
2∫x dx
=x2ln(x)
2−x2
4+C
where Cis the constant of integration.
Therefore, ∫xln(x)dx =x2ln(x)
2−x2
4+C.
Question 6
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will use the
formula: ∫u dv =uv −∫v du.
Step 1: Choose uand dv. Let u= ln(x)and dv =x dx.
Step 2: Calculate du and v. Calculate du by taking the derivative of uwith
respect to x:
du =1
xdx
Calculate vby integrating dv with respect to x:
v=1
2x2
Step 3: Apply the integration by parts formula. Using the formula ∫u dv =
uv −∫v du, we have:
∫xln(x)dx = (ln(x))( 1
2x2)−∫1
2x2·1
xdx
Step 4: Simplify and evaluate the integral. Simplify the expression:
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
5
Integrating the last term:
∫x dx =1
2x2
Step 5: Combine the terms. Putting it all together, we have:
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
where Cis the constant of integration.
Question 7
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xln(x)dx using integration by parts, we will
apply the formula:
∫u dv =uv −∫v du
where we choose u= ln(x)and dv =x dx.
Step 1: Find du and v.
du =1
xdx
v=1
2x2
Step 2: Apply integration by parts formula.
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
Step 3: Evaluate the integral.
∫xln(x)dx =1
2x2ln(x)−1
2·1
2x2+C
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
6
Question 8
Question
Evaluate the integral ∫exsin x dx using integration by parts.
Solution
To evaluate ∫exsin x dx using integration by parts, we will choose u=exand
dv = sin x dx.
Step 1: Calculate du and v.
Let u=ex(Choose u=ex)
⇒du =exdx (Calculate du)
Let dv = sin x dx (Choose dv = sin x dx)
⇒v=−cos x(Integrate dv)
Step 2: Apply the integration by parts formula ∫u dv =uv −∫v du.
∫exsin x dx =ex(−cos x)−∫(−cos x)(exdx)
=−excos x+∫excos x dx
Step 3: Apply integration by parts again to evaluate ∫excos x dx. Let
u=exand dv = cos x dx.
Step 4: Calculate du and v.
Let u=ex(Choose u=ex)
⇒du =exdx (Calculate du)
Let dv = cos x dx (Choose dv = cos x dx)
⇒v= sin x(Integrate dv)
Step 5: Apply the integration by parts formula again.
∫excos x dx =ex(sin x)−∫(sin x)(exdx)
=exsin x−∫exsin x dx
Step 6: Substitute the result back into the original integral.
∫exsin x dx =−excos x+exsin x−∫exsin x dx
7
Step 7: Solve for ∫exsin x dx.
2∫exsin x dx =−excos x+exsin x
∫exsin x dx =1
2(exsin x−excos x) + C
Therefore, ∫exsin x dx =1
2(exsin x−excos x) + C.
Question 9
Question
Evaluate the definite integral ∫π
4
0xcos(x)dx using integration by parts.
Solution
To solve the given integral, we will use integration by parts. Integration by
parts formula is given by:
∫u dv =uv −∫v du
where uand vare differentiable functions of x. In this case, we will let u=x
and dv = cos(x)dx. Then, we will differentiate uto find du and integrate dv to
find v.
Step 1: Find du and v
u=x, dv = cos(x)dx
Taking the differential of ugives:
du =dx
Integrating dv gives:
v=∫cos(x)dx = sin(x)
Step 2: Apply the integration by parts formula Using the formula ∫u dv =
uv −∫v du, we have:
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C
8
Step 3: Evaluate the definite integral Now, we can evaluate the definite
integral ∫π
4
0xcos(x)dx using the antiderivative we found:
∫π
4
0
xcos(x)dx = [xsin(x) + cos(x)]
π
4
0
=(π
4sin (π
4)+ cos (π
4))−(0 sin(0) + cos(0))
=(π
4·
√2
2+√2
2)−(0 + 1)
=π√2
8+√2
2−1
Therefore, ∫π
4
0xcos(x)dx =π√2
8+√2
2−1.
Question 10
Question
Evaluate the integral ∫x2ln x dx using integration by parts.
Solution
To solve this integral, we will use integration by parts, which states:
∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let’s choose u= ln xand dv =x2dx. Then, we have:
du =1
xdx
v=∫x2dx =x3
3
Step 2: Now, we can apply the integration by parts formula:
∫x2ln x dx =x3
3ln x−∫x3
3·1
xdx
Simplify:
∫x2ln x dx =x3
3ln x−1
3∫x2dx
Step 3: Integrate the remaining term:
∫x2ln x dx =x3
3ln x−1
3·x3
3+C
9
∫x2ln x dx =x3
3ln x−x3
9+C
Therefore, the solution to the integral ∫x2ln x dx is x3
3ln x−x3
9+C, where
Cis the constant of integration.
Question 11
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To integrate the given function ∫x2exdx, we will use integration by parts, which
states: ∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Choose uand dv. Let u=x2and dv =exdx.
Step 2: Calculate du and v.
du =d
dx (x2) = 2x dx
To find v, we need to integrate dv:
v=∫exdx =ex
Step 3: Apply the integration by parts formula.
∫x2exdx =x2ex−∫2xexdx
Step 4: Integrate the remaining integral. Using integration by parts again
with u=xand dv =exdx:
du =d
dx (x) = dx
v=∫exdx =ex
Substitute these values into the formula:
∫2xexdx = 2xex−∫2exdx
Step 5: Evaluate the final integral.
∫2exdx = 2ex
10
Step 6: Combine all the terms. Putting everything together, we have:
∫x2exdx =x2ex−2xex+ 2ex+C
where Cis the constant of integration.
Question 12
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To evaluate the integral ∫x2exdx using integration by parts, we will let u=x2
and dv =exdx. Then, we will differentiate uto find du and integrate dv to find
v. Finally, we will apply the integration by parts formula ∫u dv =uv −∫v du.
Step 1: Let u=x2and dv =exdx.Step 2: Differentiate uto find du:
du = 2x dx
Step 3: Integrate dv to find v:
v=∫exdx =ex
Step 4: Apply the integration by parts formula:
∫x2exdx =x2ex−∫ex·2x dx
∫x2exdx =x2ex−2∫xexdx
Step 5: We now have a new integral ∫xexdx. Let’s use integration by parts
again on this integral.
Step 6: Let u=xand dv =exdx.Step 7: Differentiate uto find du:
du =dx
Step 8: Integrate dv to find v:
v=ex
Step 9: Apply the integration by parts formula again:
∫xexdx =xex−∫exdx
∫xexdx =xex−ex
11
Step 10: Substitute back into the previous integral:
∫x2exdx =x2ex−2(xex−ex) + C
∫x2exdx =x2ex−2xex+ 2ex+C
Answer: ∫x2exdx =x2ex−2xex+ 2ex+C
Question 13
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will let u=
ln(x)and dv =x dx.
Step 1: Compute du and v.
du =1
xdx and v=1
2x2
Step 2: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 3: Substitute u,dv,v, and du into the formula:
= ln(x)·1
2x2−∫1
2x2·1
xdx
Step 4: Simplify the integral:
=1
2x2ln(x)−1
2∫x dx
Step 5: Integrate the remaining integral:
=1
2x2ln(x)−1
2·1
2x2+C
Step 6: Simplify the final result:
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
12
Question 14
Question
Evaluate the integral ∫xln x dx using integration by parts.
Solution
To evaluate the integral ∫xln x dx using integration by parts, we will choose
u= ln xand dv =x dx. We will then differentiate uto get du and integrate dv
to get v.
u= ln x
dv =x dx
du =1
xdx
v=1
2x2
Step 1: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 2: Substitute u,dv,v, and du into the formula:
∫xln x dx =1
2x2ln x−∫1
2x2·1
xdx
=1
2x2ln x−1
2∫x dx
Step 3: Evaluate the remaining integral:
∫x dx =1
2x2+C
Step 4: Substitute back the value of the integral and simplify:
∫xln x dx =1
2x2ln x−1
2(1
2x2)+C
=1
2x2ln x−1
4x2+C
Therefore, ∫xln x dx =1
2x2ln x−1
4x2+C, where Cis the constant of
integration.
13
Question 15
Question
Evaluate the definite integral:
∫1
0
xln(x)dx
Solution
To evaluate the given integral, we will use integration by parts by choosing
u= ln(x)and dv =x dx. Then, we will use the formula:
∫u dv =uv −∫v du
Step 1: Determine du and v.
•du =1
xdx
•v=x2
2
Step 2: Apply integration by parts.
∫1
0
xln(x)dx =[x2
2ln(x)]1
0−∫1
0
x2
2·1
xdx
=[1
2ln(1) −0]−1
2∫1
0
x dx
= 0 −1
4=−1
4
Therefore, ∫1
0xln(x)dx =−1
4.
Question 16
Question
Evaluate the integral ∫excos(x)dx using integration by parts.
Solution
To evaluate the integral ∫excos(x)dx using integration by parts, we will choose
u=exand dv = cos(x)dx. Then, we will differentiate uto get du and integrate
dv to get v.
14
Step 1: Let u=exand dv = cos(x)dx. Then, du =exdx and v=
∫cos(x)dx = sin(x).
Step 2: Apply the integration by parts formula: ∫u dv =uv −∫v du.
∫excos(x)dx =exsin(x)−∫sin(x)·exdx
=exsin(x)−(−excos(x)−∫−excos(x)dx)
=exsin(x) + excos(x)−∫excos(x)dx
Step 3: Rearrange the equation to solve for ∫excos(x)dx.
2∫excos(x)dx =exsin(x) + excos(x)
∫excos(x)dx =1
2(exsin(x) + excos(x))
Therefore, the integral ∫excos(x)dx is 1
2(exsin(x) + excos(x)) + C, where
Cis the constant of integration.
Question 17
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x dx. Then, we will differentiate uto find du and integrate
dv to find v.
Step 1: Find du and v.
•u= ln(x)
•du =1
xdx
•dv =x dx ⇒v=1
2x2
Step 2: Apply the formula for integration by parts:
∫u dv =uv −∫v du
15
Step 3: Substitute u,v,du, and dv into the integration by parts formula:
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
Step 4: Simplify the right-hand side of the equation:
1
2x2ln(x)−∫1
2x dx
Step 5: Integrate the remaining integral:
1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 18
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we use the for-
mula ∫u dv =uv −∫v du.
Step 1: Identify uand dv Let u= ln(x)and dv =x dx. Then, calculate
du and v.
•du =1
xdx
• To find v, we integrate dv:
v=∫x dx =1
2x2
Step 2: Apply the formula Now, we apply the integration by parts
formula: ∫xln(x)dx =uv −∫v du
∫xln(x)dx = ln(x)·1
2x2−∫1
2x2·1
xdx
Step 3: Simplify and integrate
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
Thus, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of inte-
gration.
16
Question 19
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xln(x)dx using integration by parts, we will
choose u= ln(x)and dv =x dx, so that du =1
xdx and v=1
2x2. Now we will
apply the integration by parts formula:
∫u dv =uv −∫v du
Step 1: Let’s calculate du and v:
•du =1
xdx
•v=1
2x2
Step 2: Now we can apply the integration by parts formula:
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
2·1
2x2+C
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 20
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx, we will use integration by parts. Recall
the formula for integration by parts:
∫u dv =uv −∫v du
17
Step 1: Let’s choose uand dv. Let u= ln(x)and dv =x dx. Then, we
have:
du =1
xdx
v=1
2x2
Step 2: Now, let’s apply the integration by parts formula:
∫xln(x)dx =uv −∫v du
= ln(x)·1
2x2−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 21
Question
Evaluate the integral ∫xsin(3x)dx using integration by parts.
Solution
To evaluate the integral ∫xsin(3x)dx, we will use integration by parts. The
formula for integration by parts is given by:
∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Identify uand dv
Let u=xand dv = sin(3x)dx. Then, calculate du and v:
du =dx
v=−1
3cos(3x)
Step 2: Apply integration by parts formula
Using the integration by parts formula, we have:
∫xsin(3x)dx =uv −∫v du
18
Step 3: Calculate the integral
=x(−1
3cos(3x))−∫(−1
3cos(3x))dx
=−x
3cos(3x) + 1
3∫cos(3x)dx
Step 4: Evaluate the remaining integral
=−x
3cos(3x) + 1
3(1
3sin(3x))+C
Step 5: Final answer
Therefore, the solution to the integral ∫xsin(3x)dx is:
−x
3cos(3x) + 1
9sin(3x) + C
where Cis the constant of integration.
Question 22
Question
Evaluate the integral ∫xsin(2x)dx using integration by parts.
Solution
To evaluate the integral ∫xsin(2x)dx using integration by parts, we will set:
u=xand dv = sin(2x)dx
Taking the differentials, we get:
du =dx and v=−1
2cos(2x)
Step 1: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 2: Substitute u,dv,du, and vinto the integration by parts formula:
∫xsin(2x)dx =x(−1
2cos(2x))−∫(−1
2cos(2x))dx
=−1
2xcos(2x) + 1
2∫cos(2x)dx
19
Step 3: Integrate ∫cos(2x)dx with respect to x:
∫cos(2x)dx =1
2sin(2x) + C
Step 4: Substitute the result back into the equation:
−1
2xcos(2x) + 1
2(1
2sin(2x))+C
Therefore, the solution to the integral ∫xsin(2x)dx is −1
2xcos(2x)+1
4sin(2x)+
C, where Cis the constant of integration.
Question 23
Question
Evaluate the integral ∫exsin x dx using integration by parts.
Solution
To evaluate the integral ∫exsin x dx using integration by parts, we will choose
u= sin xand dv =exdx.
Step 1: Compute du and v.
•du =d
dx (sin x)dx = cos x dx
•v=∫exdx =ex
Step 2: Apply the integration by parts formula, ∫u dv =uv −∫v du.
∫exsin x dx = sin x·ex−∫excos x dx
Step 3: The remaining integral ∫excos x dx can be evaluated using inte-
gration by parts again. Let’s choose u= cos xand dv =exdx.
Step 4: Compute du and v.
•du =d
dx (cos x)dx =−sin x dx
•v=∫exdx =ex
Step 5: Apply the integration by parts formula once more.
∫excos x dx = cos x·ex−∫ex(−sin x)dx
= cos x·ex+∫exsin x dx
20
Step 6: Substitute the result from Step 5 back into the initial integration
by parts formula.
∫exsin x dx = sin x·ex−(cos x·ex+∫exsin x dx)
2∫exsin x dx = (sin x−cos x)·ex
∫exsin x dx =(sin x−cos x)·ex
2+C
Therefore, ∫exsin x dx =(sin x−cos x)·ex
2+C.
Question 24
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the given integral, we will use the formula for integration by parts:
∫u dv =uv −∫v du
Let’s choose u= ln(x)and dv =x2dx. Then, we have du =1
xdx and v=1
3x3.
Step 1: Apply integration by parts:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
Step 2: Evaluate the remaining integral:
∫x2dx =1
3x3+C
Step 3: Substitute back into the original integral:
∫x2ln(x)dx =1
3x3ln(x)−1
3(1
3x3)+C
=1
3x3ln(x)−1
9x3+C
Therefore, the integral ∫x2ln(x)dx evaluates to 1
3x3ln(x)−1
9x3+C, where
Cis the constant of integration.
21
Question 25
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we let u= ln(x)
and dv =x dx. Then, du =1
xdx and v=1
2x2.
Step 1: Apply integration by parts formula: ∫u dv =uv −∫v du.
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
Step 2: Integrate the remaining term.
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C
So, the integral ∫xln(x)dx evaluates to 1
2x2ln(x)−1
4x2+C, where Cis the
constant of integration.
22
Question 2
Question
Evaluate the integral ∫xsin−1(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xsin−1(x)dx using integration by parts, we will
let u= sin−1(x)and dv =x dx. Then, we have du =1
√1−x2dx and v=1
2x2.
Step 1: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 2: Substitute u,dv,du, and vinto the formula:
∫xsin−1(x)dx =1
2x2sin−1(x)−∫1
2x2·1
√1−x2dx
Step 3: Compute the new integral:
∫1
2x2·1
√1−x2dx =1
2∫x2
√1−x2dx
Step 4: Let x= sin(t), then dx = cos(t)dt
1
2∫x2
√1−x2dx =1
2∫sin2(t)
cos(t)cos(t)dt =1
2∫sin2(t)dt
Step 5: Use the trigonometric identity sin2(t) = 1
2−1
2cos(2t)to simplify
the integral: 1
2∫sin2(t)dt =1
4t−1
4sin(2t) + C
Step 6: Substitute back xfor tand combine the results:
∫xsin−1(x)dx =1
2x2sin−1(x)−1
4arcsin(x) + 1
4x√1−x2+C
Therefore, ∫xsin−1(x)dx =1
2x2sin−1(x)−1
4arcsin(x) + 1
4x√1−x2+C
where Cis the constant of integration.
Question 3
Question
Evaluate the integral ∫excos x dx using integration by parts.
2
Solution
To evaluate ∫excos x dx using integration by parts, we will apply the formula
∫u dv =uv −∫v du. Let u=exand dv = cos x dx. Then, we have du =exdx
and v=∫cos x dx = sin x.
Step 1: Apply the integration by parts formula:
∫excos x dx =exsin x−∫sin x·exdx
=exsin x−(−∫exsin x dx)
Step 2: Apply integration by parts again to evaluate ∫exsin x dx: Let
u=exand dv = sin x dx. Then, du =exdx and v=−cos x.
∫exsin x dx =−excos x−∫(−cos x)·exdx
=−excos x−(−∫excos x dx)
Step 3: Substitute ∫excos x dx back into the equation from Step 2:
∫exsin x dx =−excos x−(−exsin x+∫excos x dx)
∫exsin x dx =−excos x+exsin x−∫excos x dx
Step 4: Substitute the result back into the original integral:
∫excos x dx =exsin x−(−excos x+exsin x−∫excos x dx)
2∫excos x dx = 2exsin x−excos x
∫excos x dx =2exsin x−excos x
2
Therefore, the solution is ∫excos x dx =2exsin x−excos x
2+C, where Cis the
constant of integration.
Question 4
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
3
Solution
To evaluate the integral ∫x2ln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x2dx. Then, we will differentiate uto find du and integrate
dv to find v.
Step 1: Let u= ln(x)and dv =x2dx. Then, we have:
du =1
xdx and v=1
3x3
Step 2: Apply integration by parts formula:
∫x2ln(x)dx =uv −∫v du
= ln(x)·1
3x3−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
Step 3: Simplify the integral:
∫x2ln(x)dx =1
3x3ln(x)−1
3∫x2dx
=1
3x3ln(x)−1
3·1
3x3+C
=1
3x3ln(x)−1
9x3+C
Therefore, ∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C, where Cis the constant of
integration.
Question 5
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will let u=
ln(x)and dv =x dx.
Step 1: Compute du and v.
• Since u= ln(x), differentiate uwith respect to xto find du:
du
dx =1
x
du =1
xdx
4
• Since dv =x dx, integrate dv with respect to xto find v:
v=∫x dx =x2
2
Step 2: Apply the integration by parts formula: ∫u dv =uv −∫v du.
∫xln(x)dx = ln(x)·x2
2−∫x2
2·1
xdx
=x2ln(x)
2−1
2∫x dx
=x2ln(x)
2−x2
4+C
where Cis the constant of integration.
Therefore, ∫xln(x)dx =x2ln(x)
2−x2
4+C.
Question 6
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will use the
formula: ∫u dv =uv −∫v du.
Step 1: Choose uand dv. Let u= ln(x)and dv =x dx.
Step 2: Calculate du and v. Calculate du by taking the derivative of uwith
respect to x:
du =1
xdx
Calculate vby integrating dv with respect to x:
v=1
2x2
Step 3: Apply the integration by parts formula. Using the formula ∫u dv =
uv −∫v du, we have:
∫xln(x)dx = (ln(x))( 1
2x2)−∫1
2x2·1
xdx
Step 4: Simplify and evaluate the integral. Simplify the expression:
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
5
Integrating the last term:
∫x dx =1
2x2
Step 5: Combine the terms. Putting it all together, we have:
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
where Cis the constant of integration.
Question 7
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xln(x)dx using integration by parts, we will
apply the formula:
∫u dv =uv −∫v du
where we choose u= ln(x)and dv =x dx.
Step 1: Find du and v.
du =1
xdx
v=1
2x2
Step 2: Apply integration by parts formula.
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
Step 3: Evaluate the integral.
∫xln(x)dx =1
2x2ln(x)−1
2·1
2x2+C
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
6
Question 8
Question
Evaluate the integral ∫exsin x dx using integration by parts.
Solution
To evaluate ∫exsin x dx using integration by parts, we will choose u=exand
dv = sin x dx.
Step 1: Calculate du and v.
Let u=ex(Choose u=ex)
⇒du =exdx (Calculate du)
Let dv = sin x dx (Choose dv = sin x dx)
⇒v=−cos x(Integrate dv)
Step 2: Apply the integration by parts formula ∫u dv =uv −∫v du.
∫exsin x dx =ex(−cos x)−∫(−cos x)(exdx)
=−excos x+∫excos x dx
Step 3: Apply integration by parts again to evaluate ∫excos x dx. Let
u=exand dv = cos x dx.
Step 4: Calculate du and v.
Let u=ex(Choose u=ex)
⇒du =exdx (Calculate du)
Let dv = cos x dx (Choose dv = cos x dx)
⇒v= sin x(Integrate dv)
Step 5: Apply the integration by parts formula again.
∫excos x dx =ex(sin x)−∫(sin x)(exdx)
=exsin x−∫exsin x dx
Step 6: Substitute the result back into the original integral.
∫exsin x dx =−excos x+exsin x−∫exsin x dx
7
Step 7: Solve for ∫exsin x dx.
2∫exsin x dx =−excos x+exsin x
∫exsin x dx =1
2(exsin x−excos x) + C
Therefore, ∫exsin x dx =1
2(exsin x−excos x) + C.
Question 9
Question
Evaluate the definite integral ∫π
4
0xcos(x)dx using integration by parts.
Solution
To solve the given integral, we will use integration by parts. Integration by
parts formula is given by:
∫u dv =uv −∫v du
where uand vare differentiable functions of x. In this case, we will let u=x
and dv = cos(x)dx. Then, we will differentiate uto find du and integrate dv to
find v.
Step 1: Find du and v
u=x, dv = cos(x)dx
Taking the differential of ugives:
du =dx
Integrating dv gives:
v=∫cos(x)dx = sin(x)
Step 2: Apply the integration by parts formula Using the formula ∫u dv =
uv −∫v du, we have:
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C
8
Step 3: Evaluate the definite integral Now, we can evaluate the definite
integral ∫π
4
0xcos(x)dx using the antiderivative we found:
∫π
4
0
xcos(x)dx = [xsin(x) + cos(x)]
π
4
0
=(π
4sin (π
4)+ cos (π
4))−(0 sin(0) + cos(0))
=(π
4·
√2
2+√2
2)−(0 + 1)
=π√2
8+√2
2−1
Therefore, ∫π
4
0xcos(x)dx =π√2
8+√2
2−1.
Question 10
Question
Evaluate the integral ∫x2ln x dx using integration by parts.
Solution
To solve this integral, we will use integration by parts, which states:
∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let’s choose u= ln xand dv =x2dx. Then, we have:
du =1
xdx
v=∫x2dx =x3
3
Step 2: Now, we can apply the integration by parts formula:
∫x2ln x dx =x3
3ln x−∫x3
3·1
xdx
Simplify:
∫x2ln x dx =x3
3ln x−1
3∫x2dx
Step 3: Integrate the remaining term:
∫x2ln x dx =x3
3ln x−1
3·x3
3+C
9
∫x2ln x dx =x3
3ln x−x3
9+C
Therefore, the solution to the integral ∫x2ln x dx is x3
3ln x−x3
9+C, where
Cis the constant of integration.
Question 11
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To integrate the given function ∫x2exdx, we will use integration by parts, which
states: ∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Choose uand dv. Let u=x2and dv =exdx.
Step 2: Calculate du and v.
du =d
dx (x2) = 2x dx
To find v, we need to integrate dv:
v=∫exdx =ex
Step 3: Apply the integration by parts formula.
∫x2exdx =x2ex−∫2xexdx
Step 4: Integrate the remaining integral. Using integration by parts again
with u=xand dv =exdx:
du =d
dx (x) = dx
v=∫exdx =ex
Substitute these values into the formula:
∫2xexdx = 2xex−∫2exdx
Step 5: Evaluate the final integral.
∫2exdx = 2ex
10
Step 6: Combine all the terms. Putting everything together, we have:
∫x2exdx =x2ex−2xex+ 2ex+C
where Cis the constant of integration.
Question 12
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To evaluate the integral ∫x2exdx using integration by parts, we will let u=x2
and dv =exdx. Then, we will differentiate uto find du and integrate dv to find
v. Finally, we will apply the integration by parts formula ∫u dv =uv −∫v du.
Step 1: Let u=x2and dv =exdx.Step 2: Differentiate uto find du:
du = 2x dx
Step 3: Integrate dv to find v:
v=∫exdx =ex
Step 4: Apply the integration by parts formula:
∫x2exdx =x2ex−∫ex·2x dx
∫x2exdx =x2ex−2∫xexdx
Step 5: We now have a new integral ∫xexdx. Let’s use integration by parts
again on this integral.
Step 6: Let u=xand dv =exdx.Step 7: Differentiate uto find du:
du =dx
Step 8: Integrate dv to find v:
v=ex
Step 9: Apply the integration by parts formula again:
∫xexdx =xex−∫exdx
∫xexdx =xex−ex
11
Step 10: Substitute back into the previous integral:
∫x2exdx =x2ex−2(xex−ex) + C
∫x2exdx =x2ex−2xex+ 2ex+C
Answer: ∫x2exdx =x2ex−2xex+ 2ex+C
Question 13
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will let u=
ln(x)and dv =x dx.
Step 1: Compute du and v.
du =1
xdx and v=1
2x2
Step 2: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 3: Substitute u,dv,v, and du into the formula:
= ln(x)·1
2x2−∫1
2x2·1
xdx
Step 4: Simplify the integral:
=1
2x2ln(x)−1
2∫x dx
Step 5: Integrate the remaining integral:
=1
2x2ln(x)−1
2·1
2x2+C
Step 6: Simplify the final result:
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
12
Question 14
Question
Evaluate the integral ∫xln x dx using integration by parts.
Solution
To evaluate the integral ∫xln x dx using integration by parts, we will choose
u= ln xand dv =x dx. We will then differentiate uto get du and integrate dv
to get v.
u= ln x
dv =x dx
du =1
xdx
v=1
2x2
Step 1: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 2: Substitute u,dv,v, and du into the formula:
∫xln x dx =1
2x2ln x−∫1
2x2·1
xdx
=1
2x2ln x−1
2∫x dx
Step 3: Evaluate the remaining integral:
∫x dx =1
2x2+C
Step 4: Substitute back the value of the integral and simplify:
∫xln x dx =1
2x2ln x−1
2(1
2x2)+C
=1
2x2ln x−1
4x2+C
Therefore, ∫xln x dx =1
2x2ln x−1
4x2+C, where Cis the constant of
integration.
13
Question 15
Question
Evaluate the definite integral:
∫1
0
xln(x)dx
Solution
To evaluate the given integral, we will use integration by parts by choosing
u= ln(x)and dv =x dx. Then, we will use the formula:
∫u dv =uv −∫v du
Step 1: Determine du and v.
•du =1
xdx
•v=x2
2
Step 2: Apply integration by parts.
∫1
0
xln(x)dx =[x2
2ln(x)]1
0−∫1
0
x2
2·1
xdx
=[1
2ln(1) −0]−1
2∫1
0
x dx
= 0 −1
4=−1
4
Therefore, ∫1
0xln(x)dx =−1
4.
Question 16
Question
Evaluate the integral ∫excos(x)dx using integration by parts.
Solution
To evaluate the integral ∫excos(x)dx using integration by parts, we will choose
u=exand dv = cos(x)dx. Then, we will differentiate uto get du and integrate
dv to get v.
14
Step 1: Let u=exand dv = cos(x)dx. Then, du =exdx and v=
∫cos(x)dx = sin(x).
Step 2: Apply the integration by parts formula: ∫u dv =uv −∫v du.
∫excos(x)dx =exsin(x)−∫sin(x)·exdx
=exsin(x)−(−excos(x)−∫−excos(x)dx)
=exsin(x) + excos(x)−∫excos(x)dx
Step 3: Rearrange the equation to solve for ∫excos(x)dx.
2∫excos(x)dx =exsin(x) + excos(x)
∫excos(x)dx =1
2(exsin(x) + excos(x))
Therefore, the integral ∫excos(x)dx is 1
2(exsin(x) + excos(x)) + C, where
Cis the constant of integration.
Question 17
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x dx. Then, we will differentiate uto find du and integrate
dv to find v.
Step 1: Find du and v.
•u= ln(x)
•du =1
xdx
•dv =x dx ⇒v=1
2x2
Step 2: Apply the formula for integration by parts:
∫u dv =uv −∫v du
15
Step 3: Substitute u,v,du, and dv into the integration by parts formula:
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
Step 4: Simplify the right-hand side of the equation:
1
2x2ln(x)−∫1
2x dx
Step 5: Integrate the remaining integral:
1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 18
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we use the for-
mula ∫u dv =uv −∫v du.
Step 1: Identify uand dv Let u= ln(x)and dv =x dx. Then, calculate
du and v.
•du =1
xdx
• To find v, we integrate dv:
v=∫x dx =1
2x2
Step 2: Apply the formula Now, we apply the integration by parts
formula: ∫xln(x)dx =uv −∫v du
∫xln(x)dx = ln(x)·1
2x2−∫1
2x2·1
xdx
Step 3: Simplify and integrate
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
Thus, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of inte-
gration.
16
Question 19
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xln(x)dx using integration by parts, we will
choose u= ln(x)and dv =x dx, so that du =1
xdx and v=1
2x2. Now we will
apply the integration by parts formula:
∫u dv =uv −∫v du
Step 1: Let’s calculate du and v:
•du =1
xdx
•v=1
2x2
Step 2: Now we can apply the integration by parts formula:
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
2·1
2x2+C
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 20
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx, we will use integration by parts. Recall
the formula for integration by parts:
∫u dv =uv −∫v du
17
Step 1: Let’s choose uand dv. Let u= ln(x)and dv =x dx. Then, we
have:
du =1
xdx
v=1
2x2
Step 2: Now, let’s apply the integration by parts formula:
∫xln(x)dx =uv −∫v du
= ln(x)·1
2x2−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 21
Question
Evaluate the integral ∫xsin(3x)dx using integration by parts.
Solution
To evaluate the integral ∫xsin(3x)dx, we will use integration by parts. The
formula for integration by parts is given by:
∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Identify uand dv
Let u=xand dv = sin(3x)dx. Then, calculate du and v:
du =dx
v=−1
3cos(3x)
Step 2: Apply integration by parts formula
Using the integration by parts formula, we have:
∫xsin(3x)dx =uv −∫v du
18
Step 3: Calculate the integral
=x(−1
3cos(3x))−∫(−1
3cos(3x))dx
=−x
3cos(3x) + 1
3∫cos(3x)dx
Step 4: Evaluate the remaining integral
=−x
3cos(3x) + 1
3(1
3sin(3x))+C
Step 5: Final answer
Therefore, the solution to the integral ∫xsin(3x)dx is:
−x
3cos(3x) + 1
9sin(3x) + C
where Cis the constant of integration.
Question 22
Question
Evaluate the integral ∫xsin(2x)dx using integration by parts.
Solution
To evaluate the integral ∫xsin(2x)dx using integration by parts, we will set:
u=xand dv = sin(2x)dx
Taking the differentials, we get:
du =dx and v=−1
2cos(2x)
Step 1: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 2: Substitute u,dv,du, and vinto the integration by parts formula:
∫xsin(2x)dx =x(−1
2cos(2x))−∫(−1
2cos(2x))dx
=−1
2xcos(2x) + 1
2∫cos(2x)dx
19
Step 3: Integrate ∫cos(2x)dx with respect to x:
∫cos(2x)dx =1
2sin(2x) + C
Step 4: Substitute the result back into the equation:
−1
2xcos(2x) + 1
2(1
2sin(2x))+C
Therefore, the solution to the integral ∫xsin(2x)dx is −1
2xcos(2x)+1
4sin(2x)+
C, where Cis the constant of integration.
Question 23
Question
Evaluate the integral ∫exsin x dx using integration by parts.
Solution
To evaluate the integral ∫exsin x dx using integration by parts, we will choose
u= sin xand dv =exdx.
Step 1: Compute du and v.
•du =d
dx (sin x)dx = cos x dx
•v=∫exdx =ex
Step 2: Apply the integration by parts formula, ∫u dv =uv −∫v du.
∫exsin x dx = sin x·ex−∫excos x dx
Step 3: The remaining integral ∫excos x dx can be evaluated using inte-
gration by parts again. Let’s choose u= cos xand dv =exdx.
Step 4: Compute du and v.
•du =d
dx (cos x)dx =−sin x dx
•v=∫exdx =ex
Step 5: Apply the integration by parts formula once more.
∫excos x dx = cos x·ex−∫ex(−sin x)dx
= cos x·ex+∫exsin x dx
20
Step 6: Substitute the result from Step 5 back into the initial integration
by parts formula.
∫exsin x dx = sin x·ex−(cos x·ex+∫exsin x dx)
2∫exsin x dx = (sin x−cos x)·ex
∫exsin x dx =(sin x−cos x)·ex
2+C
Therefore, ∫exsin x dx =(sin x−cos x)·ex
2+C.
Question 24
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the given integral, we will use the formula for integration by parts:
∫u dv =uv −∫v du
Let’s choose u= ln(x)and dv =x2dx. Then, we have du =1
xdx and v=1
3x3.
Step 1: Apply integration by parts:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
Step 2: Evaluate the remaining integral:
∫x2dx =1
3x3+C
Step 3: Substitute back into the original integral:
∫x2ln(x)dx =1
3x3ln(x)−1
3(1
3x3)+C
=1
3x3ln(x)−1
9x3+C
Therefore, the integral ∫x2ln(x)dx evaluates to 1
3x3ln(x)−1
9x3+C, where
Cis the constant of integration.
21
Question 25
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we let u= ln(x)
and dv =x dx. Then, du =1
xdx and v=1
2x2.
Step 1: Apply integration by parts formula: ∫u dv =uv −∫v du.
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
Step 2: Integrate the remaining term.
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C
So, the integral ∫xln(x)dx evaluates to 1
2x2ln(x)−1
4x2+C, where Cis the
constant of integration.
22
Question 2
Question
Evaluate the integral ∫xsin−1(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xsin−1(x)dx using integration by parts, we will
let u= sin−1(x)and dv =x dx. Then, we have du =1
√1−x2dx and v=1
2x2.
Step 1: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 2: Substitute u,dv,du, and vinto the formula:
∫xsin−1(x)dx =1
2x2sin−1(x)−∫1
2x2·1
√1−x2dx
Step 3: Compute the new integral:
∫1
2x2·1
√1−x2dx =1
2∫x2
√1−x2dx
Step 4: Let x= sin(t), then dx = cos(t)dt
1
2∫x2
√1−x2dx =1
2∫sin2(t)
cos(t)cos(t)dt =1
2∫sin2(t)dt
Step 5: Use the trigonometric identity sin2(t) = 1
2−1
2cos(2t)to simplify
the integral: 1
2∫sin2(t)dt =1
4t−1
4sin(2t) + C
Step 6: Substitute back xfor tand combine the results:
∫xsin−1(x)dx =1
2x2sin−1(x)−1
4arcsin(x) + 1
4x√1−x2+C
Therefore, ∫xsin−1(x)dx =1
2x2sin−1(x)−1
4arcsin(x) + 1
4x√1−x2+C
where Cis the constant of integration.
Question 3
Question
Evaluate the integral ∫excos x dx using integration by parts.
2
Solution
To evaluate ∫excos x dx using integration by parts, we will apply the formula
∫u dv =uv −∫v du. Let u=exand dv = cos x dx. Then, we have du =exdx
and v=∫cos x dx = sin x.
Step 1: Apply the integration by parts formula:
∫excos x dx =exsin x−∫sin x·exdx
=exsin x−(−∫exsin x dx)
Step 2: Apply integration by parts again to evaluate ∫exsin x dx: Let
u=exand dv = sin x dx. Then, du =exdx and v=−cos x.
∫exsin x dx =−excos x−∫(−cos x)·exdx
=−excos x−(−∫excos x dx)
Step 3: Substitute ∫excos x dx back into the equation from Step 2:
∫exsin x dx =−excos x−(−exsin x+∫excos x dx)
∫exsin x dx =−excos x+exsin x−∫excos x dx
Step 4: Substitute the result back into the original integral:
∫excos x dx =exsin x−(−excos x+exsin x−∫excos x dx)
2∫excos x dx = 2exsin x−excos x
∫excos x dx =2exsin x−excos x
2
Therefore, the solution is ∫excos x dx =2exsin x−excos x
2+C, where Cis the
constant of integration.
Question 4
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
3
Solution
To evaluate the integral ∫x2ln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x2dx. Then, we will differentiate uto find du and integrate
dv to find v.
Step 1: Let u= ln(x)and dv =x2dx. Then, we have:
du =1
xdx and v=1
3x3
Step 2: Apply integration by parts formula:
∫x2ln(x)dx =uv −∫v du
= ln(x)·1
3x3−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
Step 3: Simplify the integral:
∫x2ln(x)dx =1
3x3ln(x)−1
3∫x2dx
=1
3x3ln(x)−1
3·1
3x3+C
=1
3x3ln(x)−1
9x3+C
Therefore, ∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C, where Cis the constant of
integration.
Question 5
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will let u=
ln(x)and dv =x dx.
Step 1: Compute du and v.
• Since u= ln(x), differentiate uwith respect to xto find du:
du
dx =1
x
du =1
xdx
4
• Since dv =x dx, integrate dv with respect to xto find v:
v=∫x dx =x2
2
Step 2: Apply the integration by parts formula: ∫u dv =uv −∫v du.
∫xln(x)dx = ln(x)·x2
2−∫x2
2·1
xdx
=x2ln(x)
2−1
2∫x dx
=x2ln(x)
2−x2
4+C
where Cis the constant of integration.
Therefore, ∫xln(x)dx =x2ln(x)
2−x2
4+C.
Question 6
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will use the
formula: ∫u dv =uv −∫v du.
Step 1: Choose uand dv. Let u= ln(x)and dv =x dx.
Step 2: Calculate du and v. Calculate du by taking the derivative of uwith
respect to x:
du =1
xdx
Calculate vby integrating dv with respect to x:
v=1
2x2
Step 3: Apply the integration by parts formula. Using the formula ∫u dv =
uv −∫v du, we have:
∫xln(x)dx = (ln(x))( 1
2x2)−∫1
2x2·1
xdx
Step 4: Simplify and evaluate the integral. Simplify the expression:
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
5
Integrating the last term:
∫x dx =1
2x2
Step 5: Combine the terms. Putting it all together, we have:
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
where Cis the constant of integration.
Question 7
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xln(x)dx using integration by parts, we will
apply the formula:
∫u dv =uv −∫v du
where we choose u= ln(x)and dv =x dx.
Step 1: Find du and v.
du =1
xdx
v=1
2x2
Step 2: Apply integration by parts formula.
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
Step 3: Evaluate the integral.
∫xln(x)dx =1
2x2ln(x)−1
2·1
2x2+C
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
6
Question 8
Question
Evaluate the integral ∫exsin x dx using integration by parts.
Solution
To evaluate ∫exsin x dx using integration by parts, we will choose u=exand
dv = sin x dx.
Step 1: Calculate du and v.
Let u=ex(Choose u=ex)
⇒du =exdx (Calculate du)
Let dv = sin x dx (Choose dv = sin x dx)
⇒v=−cos x(Integrate dv)
Step 2: Apply the integration by parts formula ∫u dv =uv −∫v du.
∫exsin x dx =ex(−cos x)−∫(−cos x)(exdx)
=−excos x+∫excos x dx
Step 3: Apply integration by parts again to evaluate ∫excos x dx. Let
u=exand dv = cos x dx.
Step 4: Calculate du and v.
Let u=ex(Choose u=ex)
⇒du =exdx (Calculate du)
Let dv = cos x dx (Choose dv = cos x dx)
⇒v= sin x(Integrate dv)
Step 5: Apply the integration by parts formula again.
∫excos x dx =ex(sin x)−∫(sin x)(exdx)
=exsin x−∫exsin x dx
Step 6: Substitute the result back into the original integral.
∫exsin x dx =−excos x+exsin x−∫exsin x dx
7
Step 7: Solve for ∫exsin x dx.
2∫exsin x dx =−excos x+exsin x
∫exsin x dx =1
2(exsin x−excos x) + C
Therefore, ∫exsin x dx =1
2(exsin x−excos x) + C.
Question 9
Question
Evaluate the definite integral ∫π
4
0xcos(x)dx using integration by parts.
Solution
To solve the given integral, we will use integration by parts. Integration by
parts formula is given by:
∫u dv =uv −∫v du
where uand vare differentiable functions of x. In this case, we will let u=x
and dv = cos(x)dx. Then, we will differentiate uto find du and integrate dv to
find v.
Step 1: Find du and v
u=x, dv = cos(x)dx
Taking the differential of ugives:
du =dx
Integrating dv gives:
v=∫cos(x)dx = sin(x)
Step 2: Apply the integration by parts formula Using the formula ∫u dv =
uv −∫v du, we have:
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C
8
Step 3: Evaluate the definite integral Now, we can evaluate the definite
integral ∫π
4
0xcos(x)dx using the antiderivative we found:
∫π
4
0
xcos(x)dx = [xsin(x) + cos(x)]
π
4
0
=(π
4sin (π
4)+ cos (π
4))−(0 sin(0) + cos(0))
=(π
4·
√2
2+√2
2)−(0 + 1)
=π√2
8+√2
2−1
Therefore, ∫π
4
0xcos(x)dx =π√2
8+√2
2−1.
Question 10
Question
Evaluate the integral ∫x2ln x dx using integration by parts.
Solution
To solve this integral, we will use integration by parts, which states:
∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let’s choose u= ln xand dv =x2dx. Then, we have:
du =1
xdx
v=∫x2dx =x3
3
Step 2: Now, we can apply the integration by parts formula:
∫x2ln x dx =x3
3ln x−∫x3
3·1
xdx
Simplify:
∫x2ln x dx =x3
3ln x−1
3∫x2dx
Step 3: Integrate the remaining term:
∫x2ln x dx =x3
3ln x−1
3·x3
3+C
9
∫x2ln x dx =x3
3ln x−x3
9+C
Therefore, the solution to the integral ∫x2ln x dx is x3
3ln x−x3
9+C, where
Cis the constant of integration.
Question 11
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To integrate the given function ∫x2exdx, we will use integration by parts, which
states: ∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Choose uand dv. Let u=x2and dv =exdx.
Step 2: Calculate du and v.
du =d
dx (x2) = 2x dx
To find v, we need to integrate dv:
v=∫exdx =ex
Step 3: Apply the integration by parts formula.
∫x2exdx =x2ex−∫2xexdx
Step 4: Integrate the remaining integral. Using integration by parts again
with u=xand dv =exdx:
du =d
dx (x) = dx
v=∫exdx =ex
Substitute these values into the formula:
∫2xexdx = 2xex−∫2exdx
Step 5: Evaluate the final integral.
∫2exdx = 2ex
10
Step 6: Combine all the terms. Putting everything together, we have:
∫x2exdx =x2ex−2xex+ 2ex+C
where Cis the constant of integration.
Question 12
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To evaluate the integral ∫x2exdx using integration by parts, we will let u=x2
and dv =exdx. Then, we will differentiate uto find du and integrate dv to find
v. Finally, we will apply the integration by parts formula ∫u dv =uv −∫v du.
Step 1: Let u=x2and dv =exdx.Step 2: Differentiate uto find du:
du = 2x dx
Step 3: Integrate dv to find v:
v=∫exdx =ex
Step 4: Apply the integration by parts formula:
∫x2exdx =x2ex−∫ex·2x dx
∫x2exdx =x2ex−2∫xexdx
Step 5: We now have a new integral ∫xexdx. Let’s use integration by parts
again on this integral.
Step 6: Let u=xand dv =exdx.Step 7: Differentiate uto find du:
du =dx
Step 8: Integrate dv to find v:
v=ex
Step 9: Apply the integration by parts formula again:
∫xexdx =xex−∫exdx
∫xexdx =xex−ex
11
Step 10: Substitute back into the previous integral:
∫x2exdx =x2ex−2(xex−ex) + C
∫x2exdx =x2ex−2xex+ 2ex+C
Answer: ∫x2exdx =x2ex−2xex+ 2ex+C
Question 13
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will let u=
ln(x)and dv =x dx.
Step 1: Compute du and v.
du =1
xdx and v=1
2x2
Step 2: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 3: Substitute u,dv,v, and du into the formula:
= ln(x)·1
2x2−∫1
2x2·1
xdx
Step 4: Simplify the integral:
=1
2x2ln(x)−1
2∫x dx
Step 5: Integrate the remaining integral:
=1
2x2ln(x)−1
2·1
2x2+C
Step 6: Simplify the final result:
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
12
Question 14
Question
Evaluate the integral ∫xln x dx using integration by parts.
Solution
To evaluate the integral ∫xln x dx using integration by parts, we will choose
u= ln xand dv =x dx. We will then differentiate uto get du and integrate dv
to get v.
u= ln x
dv =x dx
du =1
xdx
v=1
2x2
Step 1: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 2: Substitute u,dv,v, and du into the formula:
∫xln x dx =1
2x2ln x−∫1
2x2·1
xdx
=1
2x2ln x−1
2∫x dx
Step 3: Evaluate the remaining integral:
∫x dx =1
2x2+C
Step 4: Substitute back the value of the integral and simplify:
∫xln x dx =1
2x2ln x−1
2(1
2x2)+C
=1
2x2ln x−1
4x2+C
Therefore, ∫xln x dx =1
2x2ln x−1
4x2+C, where Cis the constant of
integration.
13
Question 15
Question
Evaluate the definite integral:
∫1
0
xln(x)dx
Solution
To evaluate the given integral, we will use integration by parts by choosing
u= ln(x)and dv =x dx. Then, we will use the formula:
∫u dv =uv −∫v du
Step 1: Determine du and v.
•du =1
xdx
•v=x2
2
Step 2: Apply integration by parts.
∫1
0
xln(x)dx =[x2
2ln(x)]1
0−∫1
0
x2
2·1
xdx
=[1
2ln(1) −0]−1
2∫1
0
x dx
= 0 −1
4=−1
4
Therefore, ∫1
0xln(x)dx =−1
4.
Question 16
Question
Evaluate the integral ∫excos(x)dx using integration by parts.
Solution
To evaluate the integral ∫excos(x)dx using integration by parts, we will choose
u=exand dv = cos(x)dx. Then, we will differentiate uto get du and integrate
dv to get v.
14
Step 1: Let u=exand dv = cos(x)dx. Then, du =exdx and v=
∫cos(x)dx = sin(x).
Step 2: Apply the integration by parts formula: ∫u dv =uv −∫v du.
∫excos(x)dx =exsin(x)−∫sin(x)·exdx
=exsin(x)−(−excos(x)−∫−excos(x)dx)
=exsin(x) + excos(x)−∫excos(x)dx
Step 3: Rearrange the equation to solve for ∫excos(x)dx.
2∫excos(x)dx =exsin(x) + excos(x)
∫excos(x)dx =1
2(exsin(x) + excos(x))
Therefore, the integral ∫excos(x)dx is 1
2(exsin(x) + excos(x)) + C, where
Cis the constant of integration.
Question 17
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x dx. Then, we will differentiate uto find du and integrate
dv to find v.
Step 1: Find du and v.
•u= ln(x)
•du =1
xdx
•dv =x dx ⇒v=1
2x2
Step 2: Apply the formula for integration by parts:
∫u dv =uv −∫v du
15
Step 3: Substitute u,v,du, and dv into the integration by parts formula:
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
Step 4: Simplify the right-hand side of the equation:
1
2x2ln(x)−∫1
2x dx
Step 5: Integrate the remaining integral:
1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 18
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we use the for-
mula ∫u dv =uv −∫v du.
Step 1: Identify uand dv Let u= ln(x)and dv =x dx. Then, calculate
du and v.
•du =1
xdx
• To find v, we integrate dv:
v=∫x dx =1
2x2
Step 2: Apply the formula Now, we apply the integration by parts
formula: ∫xln(x)dx =uv −∫v du
∫xln(x)dx = ln(x)·1
2x2−∫1
2x2·1
xdx
Step 3: Simplify and integrate
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
Thus, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of inte-
gration.
16
Question 19
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xln(x)dx using integration by parts, we will
choose u= ln(x)and dv =x dx, so that du =1
xdx and v=1
2x2. Now we will
apply the integration by parts formula:
∫u dv =uv −∫v du
Step 1: Let’s calculate du and v:
•du =1
xdx
•v=1
2x2
Step 2: Now we can apply the integration by parts formula:
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
2·1
2x2+C
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 20
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx, we will use integration by parts. Recall
the formula for integration by parts:
∫u dv =uv −∫v du
17
Step 1: Let’s choose uand dv. Let u= ln(x)and dv =x dx. Then, we
have:
du =1
xdx
v=1
2x2
Step 2: Now, let’s apply the integration by parts formula:
∫xln(x)dx =uv −∫v du
= ln(x)·1
2x2−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 21
Question
Evaluate the integral ∫xsin(3x)dx using integration by parts.
Solution
To evaluate the integral ∫xsin(3x)dx, we will use integration by parts. The
formula for integration by parts is given by:
∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Identify uand dv
Let u=xand dv = sin(3x)dx. Then, calculate du and v:
du =dx
v=−1
3cos(3x)
Step 2: Apply integration by parts formula
Using the integration by parts formula, we have:
∫xsin(3x)dx =uv −∫v du
18
Step 3: Calculate the integral
=x(−1
3cos(3x))−∫(−1
3cos(3x))dx
=−x
3cos(3x) + 1
3∫cos(3x)dx
Step 4: Evaluate the remaining integral
=−x
3cos(3x) + 1
3(1
3sin(3x))+C
Step 5: Final answer
Therefore, the solution to the integral ∫xsin(3x)dx is:
−x
3cos(3x) + 1
9sin(3x) + C
where Cis the constant of integration.
Question 22
Question
Evaluate the integral ∫xsin(2x)dx using integration by parts.
Solution
To evaluate the integral ∫xsin(2x)dx using integration by parts, we will set:
u=xand dv = sin(2x)dx
Taking the differentials, we get:
du =dx and v=−1
2cos(2x)
Step 1: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 2: Substitute u,dv,du, and vinto the integration by parts formula:
∫xsin(2x)dx =x(−1
2cos(2x))−∫(−1
2cos(2x))dx
=−1
2xcos(2x) + 1
2∫cos(2x)dx
19
Step 3: Integrate ∫cos(2x)dx with respect to x:
∫cos(2x)dx =1
2sin(2x) + C
Step 4: Substitute the result back into the equation:
−1
2xcos(2x) + 1
2(1
2sin(2x))+C
Therefore, the solution to the integral ∫xsin(2x)dx is −1
2xcos(2x)+1
4sin(2x)+
C, where Cis the constant of integration.
Question 23
Question
Evaluate the integral ∫exsin x dx using integration by parts.
Solution
To evaluate the integral ∫exsin x dx using integration by parts, we will choose
u= sin xand dv =exdx.
Step 1: Compute du and v.
•du =d
dx (sin x)dx = cos x dx
•v=∫exdx =ex
Step 2: Apply the integration by parts formula, ∫u dv =uv −∫v du.
∫exsin x dx = sin x·ex−∫excos x dx
Step 3: The remaining integral ∫excos x dx can be evaluated using inte-
gration by parts again. Let’s choose u= cos xand dv =exdx.
Step 4: Compute du and v.
•du =d
dx (cos x)dx =−sin x dx
•v=∫exdx =ex
Step 5: Apply the integration by parts formula once more.
∫excos x dx = cos x·ex−∫ex(−sin x)dx
= cos x·ex+∫exsin x dx
20
Step 6: Substitute the result from Step 5 back into the initial integration
by parts formula.
∫exsin x dx = sin x·ex−(cos x·ex+∫exsin x dx)
2∫exsin x dx = (sin x−cos x)·ex
∫exsin x dx =(sin x−cos x)·ex
2+C
Therefore, ∫exsin x dx =(sin x−cos x)·ex
2+C.
Question 24
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the given integral, we will use the formula for integration by parts:
∫u dv =uv −∫v du
Let’s choose u= ln(x)and dv =x2dx. Then, we have du =1
xdx and v=1
3x3.
Step 1: Apply integration by parts:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
Step 2: Evaluate the remaining integral:
∫x2dx =1
3x3+C
Step 3: Substitute back into the original integral:
∫x2ln(x)dx =1
3x3ln(x)−1
3(1
3x3)+C
=1
3x3ln(x)−1
9x3+C
Therefore, the integral ∫x2ln(x)dx evaluates to 1
3x3ln(x)−1
9x3+C, where
Cis the constant of integration.
21
Question 25
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we let u= ln(x)
and dv =x dx. Then, du =1
xdx and v=1
2x2.
Step 1: Apply integration by parts formula: ∫u dv =uv −∫v du.
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
Step 2: Integrate the remaining term.
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C
So, the integral ∫xln(x)dx evaluates to 1
2x2ln(x)−1
4x2+C, where Cis the
constant of integration.
22
Question 2
Question
Evaluate the integral ∫xsin−1(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xsin−1(x)dx using integration by parts, we will
let u= sin−1(x)and dv =x dx. Then, we have du =1
√1−x2dx and v=1
2x2.
Step 1: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 2: Substitute u,dv,du, and vinto the formula:
∫xsin−1(x)dx =1
2x2sin−1(x)−∫1
2x2·1
√1−x2dx
Step 3: Compute the new integral:
∫1
2x2·1
√1−x2dx =1
2∫x2
√1−x2dx
Step 4: Let x= sin(t), then dx = cos(t)dt
1
2∫x2
√1−x2dx =1
2∫sin2(t)
cos(t)cos(t)dt =1
2∫sin2(t)dt
Step 5: Use the trigonometric identity sin2(t) = 1
2−1
2cos(2t)to simplify
the integral: 1
2∫sin2(t)dt =1
4t−1
4sin(2t) + C
Step 6: Substitute back xfor tand combine the results:
∫xsin−1(x)dx =1
2x2sin−1(x)−1
4arcsin(x) + 1
4x√1−x2+C
Therefore, ∫xsin−1(x)dx =1
2x2sin−1(x)−1
4arcsin(x) + 1
4x√1−x2+C
where Cis the constant of integration.
Question 3
Question
Evaluate the integral ∫excos x dx using integration by parts.
2
Solution
To evaluate ∫excos x dx using integration by parts, we will apply the formula
∫u dv =uv −∫v du. Let u=exand dv = cos x dx. Then, we have du =exdx
and v=∫cos x dx = sin x.
Step 1: Apply the integration by parts formula:
∫excos x dx =exsin x−∫sin x·exdx
=exsin x−(−∫exsin x dx)
Step 2: Apply integration by parts again to evaluate ∫exsin x dx: Let
u=exand dv = sin x dx. Then, du =exdx and v=−cos x.
∫exsin x dx =−excos x−∫(−cos x)·exdx
=−excos x−(−∫excos x dx)
Step 3: Substitute ∫excos x dx back into the equation from Step 2:
∫exsin x dx =−excos x−(−exsin x+∫excos x dx)
∫exsin x dx =−excos x+exsin x−∫excos x dx
Step 4: Substitute the result back into the original integral:
∫excos x dx =exsin x−(−excos x+exsin x−∫excos x dx)
2∫excos x dx = 2exsin x−excos x
∫excos x dx =2exsin x−excos x
2
Therefore, the solution is ∫excos x dx =2exsin x−excos x
2+C, where Cis the
constant of integration.
Question 4
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
3
Solution
To evaluate the integral ∫x2ln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x2dx. Then, we will differentiate uto find du and integrate
dv to find v.
Step 1: Let u= ln(x)and dv =x2dx. Then, we have:
du =1
xdx and v=1
3x3
Step 2: Apply integration by parts formula:
∫x2ln(x)dx =uv −∫v du
= ln(x)·1
3x3−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
Step 3: Simplify the integral:
∫x2ln(x)dx =1
3x3ln(x)−1
3∫x2dx
=1
3x3ln(x)−1
3·1
3x3+C
=1
3x3ln(x)−1
9x3+C
Therefore, ∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C, where Cis the constant of
integration.
Question 5
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will let u=
ln(x)and dv =x dx.
Step 1: Compute du and v.
• Since u= ln(x), differentiate uwith respect to xto find du:
du
dx =1
x
du =1
xdx
4
• Since dv =x dx, integrate dv with respect to xto find v:
v=∫x dx =x2
2
Step 2: Apply the integration by parts formula: ∫u dv =uv −∫v du.
∫xln(x)dx = ln(x)·x2
2−∫x2
2·1
xdx
=x2ln(x)
2−1
2∫x dx
=x2ln(x)
2−x2
4+C
where Cis the constant of integration.
Therefore, ∫xln(x)dx =x2ln(x)
2−x2
4+C.
Question 6
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will use the
formula: ∫u dv =uv −∫v du.
Step 1: Choose uand dv. Let u= ln(x)and dv =x dx.
Step 2: Calculate du and v. Calculate du by taking the derivative of uwith
respect to x:
du =1
xdx
Calculate vby integrating dv with respect to x:
v=1
2x2
Step 3: Apply the integration by parts formula. Using the formula ∫u dv =
uv −∫v du, we have:
∫xln(x)dx = (ln(x))( 1
2x2)−∫1
2x2·1
xdx
Step 4: Simplify and evaluate the integral. Simplify the expression:
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
5
Integrating the last term:
∫x dx =1
2x2
Step 5: Combine the terms. Putting it all together, we have:
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
where Cis the constant of integration.
Question 7
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xln(x)dx using integration by parts, we will
apply the formula:
∫u dv =uv −∫v du
where we choose u= ln(x)and dv =x dx.
Step 1: Find du and v.
du =1
xdx
v=1
2x2
Step 2: Apply integration by parts formula.
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
Step 3: Evaluate the integral.
∫xln(x)dx =1
2x2ln(x)−1
2·1
2x2+C
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
6
Question 8
Question
Evaluate the integral ∫exsin x dx using integration by parts.
Solution
To evaluate ∫exsin x dx using integration by parts, we will choose u=exand
dv = sin x dx.
Step 1: Calculate du and v.
Let u=ex(Choose u=ex)
⇒du =exdx (Calculate du)
Let dv = sin x dx (Choose dv = sin x dx)
⇒v=−cos x(Integrate dv)
Step 2: Apply the integration by parts formula ∫u dv =uv −∫v du.
∫exsin x dx =ex(−cos x)−∫(−cos x)(exdx)
=−excos x+∫excos x dx
Step 3: Apply integration by parts again to evaluate ∫excos x dx. Let
u=exand dv = cos x dx.
Step 4: Calculate du and v.
Let u=ex(Choose u=ex)
⇒du =exdx (Calculate du)
Let dv = cos x dx (Choose dv = cos x dx)
⇒v= sin x(Integrate dv)
Step 5: Apply the integration by parts formula again.
∫excos x dx =ex(sin x)−∫(sin x)(exdx)
=exsin x−∫exsin x dx
Step 6: Substitute the result back into the original integral.
∫exsin x dx =−excos x+exsin x−∫exsin x dx
7
Step 7: Solve for ∫exsin x dx.
2∫exsin x dx =−excos x+exsin x
∫exsin x dx =1
2(exsin x−excos x) + C
Therefore, ∫exsin x dx =1
2(exsin x−excos x) + C.
Question 9
Question
Evaluate the definite integral ∫π
4
0xcos(x)dx using integration by parts.
Solution
To solve the given integral, we will use integration by parts. Integration by
parts formula is given by:
∫u dv =uv −∫v du
where uand vare differentiable functions of x. In this case, we will let u=x
and dv = cos(x)dx. Then, we will differentiate uto find du and integrate dv to
find v.
Step 1: Find du and v
u=x, dv = cos(x)dx
Taking the differential of ugives:
du =dx
Integrating dv gives:
v=∫cos(x)dx = sin(x)
Step 2: Apply the integration by parts formula Using the formula ∫u dv =
uv −∫v du, we have:
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C
8
Step 3: Evaluate the definite integral Now, we can evaluate the definite
integral ∫π
4
0xcos(x)dx using the antiderivative we found:
∫π
4
0
xcos(x)dx = [xsin(x) + cos(x)]
π
4
0
=(π
4sin (π
4)+ cos (π
4))−(0 sin(0) + cos(0))
=(π
4·
√2
2+√2
2)−(0 + 1)
=π√2
8+√2
2−1
Therefore, ∫π
4
0xcos(x)dx =π√2
8+√2
2−1.
Question 10
Question
Evaluate the integral ∫x2ln x dx using integration by parts.
Solution
To solve this integral, we will use integration by parts, which states:
∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let’s choose u= ln xand dv =x2dx. Then, we have:
du =1
xdx
v=∫x2dx =x3
3
Step 2: Now, we can apply the integration by parts formula:
∫x2ln x dx =x3
3ln x−∫x3
3·1
xdx
Simplify:
∫x2ln x dx =x3
3ln x−1
3∫x2dx
Step 3: Integrate the remaining term:
∫x2ln x dx =x3
3ln x−1
3·x3
3+C
9
∫x2ln x dx =x3
3ln x−x3
9+C
Therefore, the solution to the integral ∫x2ln x dx is x3
3ln x−x3
9+C, where
Cis the constant of integration.
Question 11
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To integrate the given function ∫x2exdx, we will use integration by parts, which
states: ∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Choose uand dv. Let u=x2and dv =exdx.
Step 2: Calculate du and v.
du =d
dx (x2) = 2x dx
To find v, we need to integrate dv:
v=∫exdx =ex
Step 3: Apply the integration by parts formula.
∫x2exdx =x2ex−∫2xexdx
Step 4: Integrate the remaining integral. Using integration by parts again
with u=xand dv =exdx:
du =d
dx (x) = dx
v=∫exdx =ex
Substitute these values into the formula:
∫2xexdx = 2xex−∫2exdx
Step 5: Evaluate the final integral.
∫2exdx = 2ex
10
Step 6: Combine all the terms. Putting everything together, we have:
∫x2exdx =x2ex−2xex+ 2ex+C
where Cis the constant of integration.
Question 12
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To evaluate the integral ∫x2exdx using integration by parts, we will let u=x2
and dv =exdx. Then, we will differentiate uto find du and integrate dv to find
v. Finally, we will apply the integration by parts formula ∫u dv =uv −∫v du.
Step 1: Let u=x2and dv =exdx.Step 2: Differentiate uto find du:
du = 2x dx
Step 3: Integrate dv to find v:
v=∫exdx =ex
Step 4: Apply the integration by parts formula:
∫x2exdx =x2ex−∫ex·2x dx
∫x2exdx =x2ex−2∫xexdx
Step 5: We now have a new integral ∫xexdx. Let’s use integration by parts
again on this integral.
Step 6: Let u=xand dv =exdx.Step 7: Differentiate uto find du:
du =dx
Step 8: Integrate dv to find v:
v=ex
Step 9: Apply the integration by parts formula again:
∫xexdx =xex−∫exdx
∫xexdx =xex−ex
11
Step 10: Substitute back into the previous integral:
∫x2exdx =x2ex−2(xex−ex) + C
∫x2exdx =x2ex−2xex+ 2ex+C
Answer: ∫x2exdx =x2ex−2xex+ 2ex+C
Question 13
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will let u=
ln(x)and dv =x dx.
Step 1: Compute du and v.
du =1
xdx and v=1
2x2
Step 2: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 3: Substitute u,dv,v, and du into the formula:
= ln(x)·1
2x2−∫1
2x2·1
xdx
Step 4: Simplify the integral:
=1
2x2ln(x)−1
2∫x dx
Step 5: Integrate the remaining integral:
=1
2x2ln(x)−1
2·1
2x2+C
Step 6: Simplify the final result:
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
12
Question 14
Question
Evaluate the integral ∫xln x dx using integration by parts.
Solution
To evaluate the integral ∫xln x dx using integration by parts, we will choose
u= ln xand dv =x dx. We will then differentiate uto get du and integrate dv
to get v.
u= ln x
dv =x dx
du =1
xdx
v=1
2x2
Step 1: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 2: Substitute u,dv,v, and du into the formula:
∫xln x dx =1
2x2ln x−∫1
2x2·1
xdx
=1
2x2ln x−1
2∫x dx
Step 3: Evaluate the remaining integral:
∫x dx =1
2x2+C
Step 4: Substitute back the value of the integral and simplify:
∫xln x dx =1
2x2ln x−1
2(1
2x2)+C
=1
2x2ln x−1
4x2+C
Therefore, ∫xln x dx =1
2x2ln x−1
4x2+C, where Cis the constant of
integration.
13
Question 15
Question
Evaluate the definite integral:
∫1
0
xln(x)dx
Solution
To evaluate the given integral, we will use integration by parts by choosing
u= ln(x)and dv =x dx. Then, we will use the formula:
∫u dv =uv −∫v du
Step 1: Determine du and v.
•du =1
xdx
•v=x2
2
Step 2: Apply integration by parts.
∫1
0
xln(x)dx =[x2
2ln(x)]1
0−∫1
0
x2
2·1
xdx
=[1
2ln(1) −0]−1
2∫1
0
x dx
= 0 −1
4=−1
4
Therefore, ∫1
0xln(x)dx =−1
4.
Question 16
Question
Evaluate the integral ∫excos(x)dx using integration by parts.
Solution
To evaluate the integral ∫excos(x)dx using integration by parts, we will choose
u=exand dv = cos(x)dx. Then, we will differentiate uto get du and integrate
dv to get v.
14
Step 1: Let u=exand dv = cos(x)dx. Then, du =exdx and v=
∫cos(x)dx = sin(x).
Step 2: Apply the integration by parts formula: ∫u dv =uv −∫v du.
∫excos(x)dx =exsin(x)−∫sin(x)·exdx
=exsin(x)−(−excos(x)−∫−excos(x)dx)
=exsin(x) + excos(x)−∫excos(x)dx
Step 3: Rearrange the equation to solve for ∫excos(x)dx.
2∫excos(x)dx =exsin(x) + excos(x)
∫excos(x)dx =1
2(exsin(x) + excos(x))
Therefore, the integral ∫excos(x)dx is 1
2(exsin(x) + excos(x)) + C, where
Cis the constant of integration.
Question 17
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x dx. Then, we will differentiate uto find du and integrate
dv to find v.
Step 1: Find du and v.
•u= ln(x)
•du =1
xdx
•dv =x dx ⇒v=1
2x2
Step 2: Apply the formula for integration by parts:
∫u dv =uv −∫v du
15
Step 3: Substitute u,v,du, and dv into the integration by parts formula:
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
Step 4: Simplify the right-hand side of the equation:
1
2x2ln(x)−∫1
2x dx
Step 5: Integrate the remaining integral:
1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 18
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we use the for-
mula ∫u dv =uv −∫v du.
Step 1: Identify uand dv Let u= ln(x)and dv =x dx. Then, calculate
du and v.
•du =1
xdx
• To find v, we integrate dv:
v=∫x dx =1
2x2
Step 2: Apply the formula Now, we apply the integration by parts
formula: ∫xln(x)dx =uv −∫v du
∫xln(x)dx = ln(x)·1
2x2−∫1
2x2·1
xdx
Step 3: Simplify and integrate
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
Thus, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of inte-
gration.
16
Question 19
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xln(x)dx using integration by parts, we will
choose u= ln(x)and dv =x dx, so that du =1
xdx and v=1
2x2. Now we will
apply the integration by parts formula:
∫u dv =uv −∫v du
Step 1: Let’s calculate du and v:
•du =1
xdx
•v=1
2x2
Step 2: Now we can apply the integration by parts formula:
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
2·1
2x2+C
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 20
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx, we will use integration by parts. Recall
the formula for integration by parts:
∫u dv =uv −∫v du
17
Step 1: Let’s choose uand dv. Let u= ln(x)and dv =x dx. Then, we
have:
du =1
xdx
v=1
2x2
Step 2: Now, let’s apply the integration by parts formula:
∫xln(x)dx =uv −∫v du
= ln(x)·1
2x2−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 21
Question
Evaluate the integral ∫xsin(3x)dx using integration by parts.
Solution
To evaluate the integral ∫xsin(3x)dx, we will use integration by parts. The
formula for integration by parts is given by:
∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Identify uand dv
Let u=xand dv = sin(3x)dx. Then, calculate du and v:
du =dx
v=−1
3cos(3x)
Step 2: Apply integration by parts formula
Using the integration by parts formula, we have:
∫xsin(3x)dx =uv −∫v du
18
Step 3: Calculate the integral
=x(−1
3cos(3x))−∫(−1
3cos(3x))dx
=−x
3cos(3x) + 1
3∫cos(3x)dx
Step 4: Evaluate the remaining integral
=−x
3cos(3x) + 1
3(1
3sin(3x))+C
Step 5: Final answer
Therefore, the solution to the integral ∫xsin(3x)dx is:
−x
3cos(3x) + 1
9sin(3x) + C
where Cis the constant of integration.
Question 22
Question
Evaluate the integral ∫xsin(2x)dx using integration by parts.
Solution
To evaluate the integral ∫xsin(2x)dx using integration by parts, we will set:
u=xand dv = sin(2x)dx
Taking the differentials, we get:
du =dx and v=−1
2cos(2x)
Step 1: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 2: Substitute u,dv,du, and vinto the integration by parts formula:
∫xsin(2x)dx =x(−1
2cos(2x))−∫(−1
2cos(2x))dx
=−1
2xcos(2x) + 1
2∫cos(2x)dx
19
Step 3: Integrate ∫cos(2x)dx with respect to x:
∫cos(2x)dx =1
2sin(2x) + C
Step 4: Substitute the result back into the equation:
−1
2xcos(2x) + 1
2(1
2sin(2x))+C
Therefore, the solution to the integral ∫xsin(2x)dx is −1
2xcos(2x)+1
4sin(2x)+
C, where Cis the constant of integration.
Question 23
Question
Evaluate the integral ∫exsin x dx using integration by parts.
Solution
To evaluate the integral ∫exsin x dx using integration by parts, we will choose
u= sin xand dv =exdx.
Step 1: Compute du and v.
•du =d
dx (sin x)dx = cos x dx
•v=∫exdx =ex
Step 2: Apply the integration by parts formula, ∫u dv =uv −∫v du.
∫exsin x dx = sin x·ex−∫excos x dx
Step 3: The remaining integral ∫excos x dx can be evaluated using inte-
gration by parts again. Let’s choose u= cos xand dv =exdx.
Step 4: Compute du and v.
•du =d
dx (cos x)dx =−sin x dx
•v=∫exdx =ex
Step 5: Apply the integration by parts formula once more.
∫excos x dx = cos x·ex−∫ex(−sin x)dx
= cos x·ex+∫exsin x dx
20
Step 6: Substitute the result from Step 5 back into the initial integration
by parts formula.
∫exsin x dx = sin x·ex−(cos x·ex+∫exsin x dx)
2∫exsin x dx = (sin x−cos x)·ex
∫exsin x dx =(sin x−cos x)·ex
2+C
Therefore, ∫exsin x dx =(sin x−cos x)·ex
2+C.
Question 24
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the given integral, we will use the formula for integration by parts:
∫u dv =uv −∫v du
Let’s choose u= ln(x)and dv =x2dx. Then, we have du =1
xdx and v=1
3x3.
Step 1: Apply integration by parts:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
Step 2: Evaluate the remaining integral:
∫x2dx =1
3x3+C
Step 3: Substitute back into the original integral:
∫x2ln(x)dx =1
3x3ln(x)−1
3(1
3x3)+C
=1
3x3ln(x)−1
9x3+C
Therefore, the integral ∫x2ln(x)dx evaluates to 1
3x3ln(x)−1
9x3+C, where
Cis the constant of integration.
21
Question 25
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we let u= ln(x)
and dv =x dx. Then, du =1
xdx and v=1
2x2.
Step 1: Apply integration by parts formula: ∫u dv =uv −∫v du.
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
Step 2: Integrate the remaining term.
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C
So, the integral ∫xln(x)dx evaluates to 1
2x2ln(x)−1
4x2+C, where Cis the
constant of integration.
22
Question 2
Question
Evaluate the integral ∫xsin−1(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xsin−1(x)dx using integration by parts, we will
let u= sin−1(x)and dv =x dx. Then, we have du =1
√1−x2dx and v=1
2x2.
Step 1: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 2: Substitute u,dv,du, and vinto the formula:
∫xsin−1(x)dx =1
2x2sin−1(x)−∫1
2x2·1
√1−x2dx
Step 3: Compute the new integral:
∫1
2x2·1
√1−x2dx =1
2∫x2
√1−x2dx
Step 4: Let x= sin(t), then dx = cos(t)dt
1
2∫x2
√1−x2dx =1
2∫sin2(t)
cos(t)cos(t)dt =1
2∫sin2(t)dt
Step 5: Use the trigonometric identity sin2(t) = 1
2−1
2cos(2t)to simplify
the integral: 1
2∫sin2(t)dt =1
4t−1
4sin(2t) + C
Step 6: Substitute back xfor tand combine the results:
∫xsin−1(x)dx =1
2x2sin−1(x)−1
4arcsin(x) + 1
4x√1−x2+C
Therefore, ∫xsin−1(x)dx =1
2x2sin−1(x)−1
4arcsin(x) + 1
4x√1−x2+C
where Cis the constant of integration.
Question 3
Question
Evaluate the integral ∫excos x dx using integration by parts.
2
Solution
To evaluate ∫excos x dx using integration by parts, we will apply the formula
∫u dv =uv −∫v du. Let u=exand dv = cos x dx. Then, we have du =exdx
and v=∫cos x dx = sin x.
Step 1: Apply the integration by parts formula:
∫excos x dx =exsin x−∫sin x·exdx
=exsin x−(−∫exsin x dx)
Step 2: Apply integration by parts again to evaluate ∫exsin x dx: Let
u=exand dv = sin x dx. Then, du =exdx and v=−cos x.
∫exsin x dx =−excos x−∫(−cos x)·exdx
=−excos x−(−∫excos x dx)
Step 3: Substitute ∫excos x dx back into the equation from Step 2:
∫exsin x dx =−excos x−(−exsin x+∫excos x dx)
∫exsin x dx =−excos x+exsin x−∫excos x dx
Step 4: Substitute the result back into the original integral:
∫excos x dx =exsin x−(−excos x+exsin x−∫excos x dx)
2∫excos x dx = 2exsin x−excos x
∫excos x dx =2exsin x−excos x
2
Therefore, the solution is ∫excos x dx =2exsin x−excos x
2+C, where Cis the
constant of integration.
Question 4
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
3
Solution
To evaluate the integral ∫x2ln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x2dx. Then, we will differentiate uto find du and integrate
dv to find v.
Step 1: Let u= ln(x)and dv =x2dx. Then, we have:
du =1
xdx and v=1
3x3
Step 2: Apply integration by parts formula:
∫x2ln(x)dx =uv −∫v du
= ln(x)·1
3x3−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
Step 3: Simplify the integral:
∫x2ln(x)dx =1
3x3ln(x)−1
3∫x2dx
=1
3x3ln(x)−1
3·1
3x3+C
=1
3x3ln(x)−1
9x3+C
Therefore, ∫x2ln(x)dx =1
3x3ln(x)−1
9x3+C, where Cis the constant of
integration.
Question 5
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will let u=
ln(x)and dv =x dx.
Step 1: Compute du and v.
• Since u= ln(x), differentiate uwith respect to xto find du:
du
dx =1
x
du =1
xdx
4
• Since dv =x dx, integrate dv with respect to xto find v:
v=∫x dx =x2
2
Step 2: Apply the integration by parts formula: ∫u dv =uv −∫v du.
∫xln(x)dx = ln(x)·x2
2−∫x2
2·1
xdx
=x2ln(x)
2−1
2∫x dx
=x2ln(x)
2−x2
4+C
where Cis the constant of integration.
Therefore, ∫xln(x)dx =x2ln(x)
2−x2
4+C.
Question 6
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will use the
formula: ∫u dv =uv −∫v du.
Step 1: Choose uand dv. Let u= ln(x)and dv =x dx.
Step 2: Calculate du and v. Calculate du by taking the derivative of uwith
respect to x:
du =1
xdx
Calculate vby integrating dv with respect to x:
v=1
2x2
Step 3: Apply the integration by parts formula. Using the formula ∫u dv =
uv −∫v du, we have:
∫xln(x)dx = (ln(x))( 1
2x2)−∫1
2x2·1
xdx
Step 4: Simplify and evaluate the integral. Simplify the expression:
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
5
Integrating the last term:
∫x dx =1
2x2
Step 5: Combine the terms. Putting it all together, we have:
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
where Cis the constant of integration.
Question 7
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xln(x)dx using integration by parts, we will
apply the formula:
∫u dv =uv −∫v du
where we choose u= ln(x)and dv =x dx.
Step 1: Find du and v.
du =1
xdx
v=1
2x2
Step 2: Apply integration by parts formula.
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
Step 3: Evaluate the integral.
∫xln(x)dx =1
2x2ln(x)−1
2·1
2x2+C
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
6
Question 8
Question
Evaluate the integral ∫exsin x dx using integration by parts.
Solution
To evaluate ∫exsin x dx using integration by parts, we will choose u=exand
dv = sin x dx.
Step 1: Calculate du and v.
Let u=ex(Choose u=ex)
⇒du =exdx (Calculate du)
Let dv = sin x dx (Choose dv = sin x dx)
⇒v=−cos x(Integrate dv)
Step 2: Apply the integration by parts formula ∫u dv =uv −∫v du.
∫exsin x dx =ex(−cos x)−∫(−cos x)(exdx)
=−excos x+∫excos x dx
Step 3: Apply integration by parts again to evaluate ∫excos x dx. Let
u=exand dv = cos x dx.
Step 4: Calculate du and v.
Let u=ex(Choose u=ex)
⇒du =exdx (Calculate du)
Let dv = cos x dx (Choose dv = cos x dx)
⇒v= sin x(Integrate dv)
Step 5: Apply the integration by parts formula again.
∫excos x dx =ex(sin x)−∫(sin x)(exdx)
=exsin x−∫exsin x dx
Step 6: Substitute the result back into the original integral.
∫exsin x dx =−excos x+exsin x−∫exsin x dx
7
Step 7: Solve for ∫exsin x dx.
2∫exsin x dx =−excos x+exsin x
∫exsin x dx =1
2(exsin x−excos x) + C
Therefore, ∫exsin x dx =1
2(exsin x−excos x) + C.
Question 9
Question
Evaluate the definite integral ∫π
4
0xcos(x)dx using integration by parts.
Solution
To solve the given integral, we will use integration by parts. Integration by
parts formula is given by:
∫u dv =uv −∫v du
where uand vare differentiable functions of x. In this case, we will let u=x
and dv = cos(x)dx. Then, we will differentiate uto find du and integrate dv to
find v.
Step 1: Find du and v
u=x, dv = cos(x)dx
Taking the differential of ugives:
du =dx
Integrating dv gives:
v=∫cos(x)dx = sin(x)
Step 2: Apply the integration by parts formula Using the formula ∫u dv =
uv −∫v du, we have:
∫xcos(x)dx =uv −∫v du
=xsin(x)−∫sin(x)dx
=xsin(x) + cos(x) + C
8
Step 3: Evaluate the definite integral Now, we can evaluate the definite
integral ∫π
4
0xcos(x)dx using the antiderivative we found:
∫π
4
0
xcos(x)dx = [xsin(x) + cos(x)]
π
4
0
=(π
4sin (π
4)+ cos (π
4))−(0 sin(0) + cos(0))
=(π
4·
√2
2+√2
2)−(0 + 1)
=π√2
8+√2
2−1
Therefore, ∫π
4
0xcos(x)dx =π√2
8+√2
2−1.
Question 10
Question
Evaluate the integral ∫x2ln x dx using integration by parts.
Solution
To solve this integral, we will use integration by parts, which states:
∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Let’s choose u= ln xand dv =x2dx. Then, we have:
du =1
xdx
v=∫x2dx =x3
3
Step 2: Now, we can apply the integration by parts formula:
∫x2ln x dx =x3
3ln x−∫x3
3·1
xdx
Simplify:
∫x2ln x dx =x3
3ln x−1
3∫x2dx
Step 3: Integrate the remaining term:
∫x2ln x dx =x3
3ln x−1
3·x3
3+C
9
∫x2ln x dx =x3
3ln x−x3
9+C
Therefore, the solution to the integral ∫x2ln x dx is x3
3ln x−x3
9+C, where
Cis the constant of integration.
Question 11
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To integrate the given function ∫x2exdx, we will use integration by parts, which
states: ∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Choose uand dv. Let u=x2and dv =exdx.
Step 2: Calculate du and v.
du =d
dx (x2) = 2x dx
To find v, we need to integrate dv:
v=∫exdx =ex
Step 3: Apply the integration by parts formula.
∫x2exdx =x2ex−∫2xexdx
Step 4: Integrate the remaining integral. Using integration by parts again
with u=xand dv =exdx:
du =d
dx (x) = dx
v=∫exdx =ex
Substitute these values into the formula:
∫2xexdx = 2xex−∫2exdx
Step 5: Evaluate the final integral.
∫2exdx = 2ex
10
Step 6: Combine all the terms. Putting everything together, we have:
∫x2exdx =x2ex−2xex+ 2ex+C
where Cis the constant of integration.
Question 12
Question
Evaluate the integral ∫x2exdx using integration by parts.
Solution
To evaluate the integral ∫x2exdx using integration by parts, we will let u=x2
and dv =exdx. Then, we will differentiate uto find du and integrate dv to find
v. Finally, we will apply the integration by parts formula ∫u dv =uv −∫v du.
Step 1: Let u=x2and dv =exdx.Step 2: Differentiate uto find du:
du = 2x dx
Step 3: Integrate dv to find v:
v=∫exdx =ex
Step 4: Apply the integration by parts formula:
∫x2exdx =x2ex−∫ex·2x dx
∫x2exdx =x2ex−2∫xexdx
Step 5: We now have a new integral ∫xexdx. Let’s use integration by parts
again on this integral.
Step 6: Let u=xand dv =exdx.Step 7: Differentiate uto find du:
du =dx
Step 8: Integrate dv to find v:
v=ex
Step 9: Apply the integration by parts formula again:
∫xexdx =xex−∫exdx
∫xexdx =xex−ex
11
Step 10: Substitute back into the previous integral:
∫x2exdx =x2ex−2(xex−ex) + C
∫x2exdx =x2ex−2xex+ 2ex+C
Answer: ∫x2exdx =x2ex−2xex+ 2ex+C
Question 13
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will let u=
ln(x)and dv =x dx.
Step 1: Compute du and v.
du =1
xdx and v=1
2x2
Step 2: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 3: Substitute u,dv,v, and du into the formula:
= ln(x)·1
2x2−∫1
2x2·1
xdx
Step 4: Simplify the integral:
=1
2x2ln(x)−1
2∫x dx
Step 5: Integrate the remaining integral:
=1
2x2ln(x)−1
2·1
2x2+C
Step 6: Simplify the final result:
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
12
Question 14
Question
Evaluate the integral ∫xln x dx using integration by parts.
Solution
To evaluate the integral ∫xln x dx using integration by parts, we will choose
u= ln xand dv =x dx. We will then differentiate uto get du and integrate dv
to get v.
u= ln x
dv =x dx
du =1
xdx
v=1
2x2
Step 1: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 2: Substitute u,dv,v, and du into the formula:
∫xln x dx =1
2x2ln x−∫1
2x2·1
xdx
=1
2x2ln x−1
2∫x dx
Step 3: Evaluate the remaining integral:
∫x dx =1
2x2+C
Step 4: Substitute back the value of the integral and simplify:
∫xln x dx =1
2x2ln x−1
2(1
2x2)+C
=1
2x2ln x−1
4x2+C
Therefore, ∫xln x dx =1
2x2ln x−1
4x2+C, where Cis the constant of
integration.
13
Question 15
Question
Evaluate the definite integral:
∫1
0
xln(x)dx
Solution
To evaluate the given integral, we will use integration by parts by choosing
u= ln(x)and dv =x dx. Then, we will use the formula:
∫u dv =uv −∫v du
Step 1: Determine du and v.
•du =1
xdx
•v=x2
2
Step 2: Apply integration by parts.
∫1
0
xln(x)dx =[x2
2ln(x)]1
0−∫1
0
x2
2·1
xdx
=[1
2ln(1) −0]−1
2∫1
0
x dx
= 0 −1
4=−1
4
Therefore, ∫1
0xln(x)dx =−1
4.
Question 16
Question
Evaluate the integral ∫excos(x)dx using integration by parts.
Solution
To evaluate the integral ∫excos(x)dx using integration by parts, we will choose
u=exand dv = cos(x)dx. Then, we will differentiate uto get du and integrate
dv to get v.
14
Step 1: Let u=exand dv = cos(x)dx. Then, du =exdx and v=
∫cos(x)dx = sin(x).
Step 2: Apply the integration by parts formula: ∫u dv =uv −∫v du.
∫excos(x)dx =exsin(x)−∫sin(x)·exdx
=exsin(x)−(−excos(x)−∫−excos(x)dx)
=exsin(x) + excos(x)−∫excos(x)dx
Step 3: Rearrange the equation to solve for ∫excos(x)dx.
2∫excos(x)dx =exsin(x) + excos(x)
∫excos(x)dx =1
2(exsin(x) + excos(x))
Therefore, the integral ∫excos(x)dx is 1
2(exsin(x) + excos(x)) + C, where
Cis the constant of integration.
Question 17
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we will choose
u= ln(x)and dv =x dx. Then, we will differentiate uto find du and integrate
dv to find v.
Step 1: Find du and v.
•u= ln(x)
•du =1
xdx
•dv =x dx ⇒v=1
2x2
Step 2: Apply the formula for integration by parts:
∫u dv =uv −∫v du
15
Step 3: Substitute u,v,du, and dv into the integration by parts formula:
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
Step 4: Simplify the right-hand side of the equation:
1
2x2ln(x)−∫1
2x dx
Step 5: Integrate the remaining integral:
1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 18
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we use the for-
mula ∫u dv =uv −∫v du.
Step 1: Identify uand dv Let u= ln(x)and dv =x dx. Then, calculate
du and v.
•du =1
xdx
• To find v, we integrate dv:
v=∫x dx =1
2x2
Step 2: Apply the formula Now, we apply the integration by parts
formula: ∫xln(x)dx =uv −∫v du
∫xln(x)dx = ln(x)·1
2x2−∫1
2x2·1
xdx
Step 3: Simplify and integrate
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
∫xln(x)dx =1
2x2ln(x)−1
4x2+C
Thus, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of inte-
gration.
16
Question 19
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the given integral ∫xln(x)dx using integration by parts, we will
choose u= ln(x)and dv =x dx, so that du =1
xdx and v=1
2x2. Now we will
apply the integration by parts formula:
∫u dv =uv −∫v du
Step 1: Let’s calculate du and v:
•du =1
xdx
•v=1
2x2
Step 2: Now we can apply the integration by parts formula:
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
2·1
2x2+C
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 20
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx, we will use integration by parts. Recall
the formula for integration by parts:
∫u dv =uv −∫v du
17
Step 1: Let’s choose uand dv. Let u= ln(x)and dv =x dx. Then, we
have:
du =1
xdx
v=1
2x2
Step 2: Now, let’s apply the integration by parts formula:
∫xln(x)dx =uv −∫v du
= ln(x)·1
2x2−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C
Therefore, ∫xln(x)dx =1
2x2ln(x)−1
4x2+C, where Cis the constant of
integration.
Question 21
Question
Evaluate the integral ∫xsin(3x)dx using integration by parts.
Solution
To evaluate the integral ∫xsin(3x)dx, we will use integration by parts. The
formula for integration by parts is given by:
∫u dv =uv −∫v du
where uand vare differentiable functions of x.
Step 1: Identify uand dv
Let u=xand dv = sin(3x)dx. Then, calculate du and v:
du =dx
v=−1
3cos(3x)
Step 2: Apply integration by parts formula
Using the integration by parts formula, we have:
∫xsin(3x)dx =uv −∫v du
18
Step 3: Calculate the integral
=x(−1
3cos(3x))−∫(−1
3cos(3x))dx
=−x
3cos(3x) + 1
3∫cos(3x)dx
Step 4: Evaluate the remaining integral
=−x
3cos(3x) + 1
3(1
3sin(3x))+C
Step 5: Final answer
Therefore, the solution to the integral ∫xsin(3x)dx is:
−x
3cos(3x) + 1
9sin(3x) + C
where Cis the constant of integration.
Question 22
Question
Evaluate the integral ∫xsin(2x)dx using integration by parts.
Solution
To evaluate the integral ∫xsin(2x)dx using integration by parts, we will set:
u=xand dv = sin(2x)dx
Taking the differentials, we get:
du =dx and v=−1
2cos(2x)
Step 1: Apply the integration by parts formula:
∫u dv =uv −∫v du
Step 2: Substitute u,dv,du, and vinto the integration by parts formula:
∫xsin(2x)dx =x(−1
2cos(2x))−∫(−1
2cos(2x))dx
=−1
2xcos(2x) + 1
2∫cos(2x)dx
19
Step 3: Integrate ∫cos(2x)dx with respect to x:
∫cos(2x)dx =1
2sin(2x) + C
Step 4: Substitute the result back into the equation:
−1
2xcos(2x) + 1
2(1
2sin(2x))+C
Therefore, the solution to the integral ∫xsin(2x)dx is −1
2xcos(2x)+1
4sin(2x)+
C, where Cis the constant of integration.
Question 23
Question
Evaluate the integral ∫exsin x dx using integration by parts.
Solution
To evaluate the integral ∫exsin x dx using integration by parts, we will choose
u= sin xand dv =exdx.
Step 1: Compute du and v.
•du =d
dx (sin x)dx = cos x dx
•v=∫exdx =ex
Step 2: Apply the integration by parts formula, ∫u dv =uv −∫v du.
∫exsin x dx = sin x·ex−∫excos x dx
Step 3: The remaining integral ∫excos x dx can be evaluated using inte-
gration by parts again. Let’s choose u= cos xand dv =exdx.
Step 4: Compute du and v.
•du =d
dx (cos x)dx =−sin x dx
•v=∫exdx =ex
Step 5: Apply the integration by parts formula once more.
∫excos x dx = cos x·ex−∫ex(−sin x)dx
= cos x·ex+∫exsin x dx
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Step 6: Substitute the result from Step 5 back into the initial integration
by parts formula.
∫exsin x dx = sin x·ex−(cos x·ex+∫exsin x dx)
2∫exsin x dx = (sin x−cos x)·ex
∫exsin x dx =(sin x−cos x)·ex
2+C
Therefore, ∫exsin x dx =(sin x−cos x)·ex
2+C.
Question 24
Question
Evaluate the integral ∫x2ln(x)dx using integration by parts.
Solution
To evaluate the given integral, we will use the formula for integration by parts:
∫u dv =uv −∫v du
Let’s choose u= ln(x)and dv =x2dx. Then, we have du =1
xdx and v=1
3x3.
Step 1: Apply integration by parts:
∫x2ln(x)dx =1
3x3ln(x)−∫1
3x3·1
xdx
=1
3x3ln(x)−1
3∫x2dx
Step 2: Evaluate the remaining integral:
∫x2dx =1
3x3+C
Step 3: Substitute back into the original integral:
∫x2ln(x)dx =1
3x3ln(x)−1
3(1
3x3)+C
=1
3x3ln(x)−1
9x3+C
Therefore, the integral ∫x2ln(x)dx evaluates to 1
3x3ln(x)−1
9x3+C, where
Cis the constant of integration.
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Question 25
Question
Evaluate the integral ∫xln(x)dx using integration by parts.
Solution
To evaluate the integral ∫xln(x)dx using integration by parts, we let u= ln(x)
and dv =x dx. Then, du =1
xdx and v=1
2x2.
Step 1: Apply integration by parts formula: ∫u dv =uv −∫v du.
∫xln(x)dx =1
2x2ln(x)−∫1
2x2·1
xdx
=1
2x2ln(x)−1
2∫x dx
Step 2: Integrate the remaining term.
∫xln(x)dx =1
2x2ln(x)−1
2∫x dx
=1
2x2ln(x)−1
4x2+C
So, the integral ∫xln(x)dx evaluates to 1
2x2ln(x)−1
4x2+C, where Cis the
constant of integration.
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