INTEGRATION PROBLEMS WITH DETAILED
SOLUTIONS, VERIFICATION, AND ALTERNATIVE
METHODS
Problem 1
Evaluate the integral:
∫𝑥2ln(𝑥) 𝑑𝑥
Solution
Using integration by parts, let 𝑢=ln(𝑥) and 𝑑𝑣=𝑥2 𝑑𝑥:
𝑑𝑢=1
𝑥 𝑑𝑥 and 𝑣=𝑥3
3
∫𝑥2ln(𝑥) 𝑑𝑥=ln(𝑥)⋅𝑥3
3−∫ 𝑥3
3⋅1
𝑥 𝑑𝑥
=𝑥3ln(𝑥)
3−1
3∫𝑥2 𝑑𝑥
=𝑥3ln(𝑥)
3−1
3⋅𝑥3
3
=𝑥3ln(𝑥)
3−𝑥3
9
=𝑥3(ln(𝑥)−1
3)
3+𝐶
Verification
Differentiate 𝐹(𝑥)=𝑥3(ln(𝑥)−1
3)
3:
𝐹′(𝑥)=1
3(3𝑥2(ln(𝑥)−1
3)+𝑥3⋅1
𝑥)
=𝑥2ln(𝑥)
Alternative Approach
Substitute 𝑥 =𝑒𝑡, 𝑑𝑥=𝑒𝑡 𝑑𝑡:
∫𝑥2ln(𝑥) 𝑑𝑥=∫𝑒3𝑡 ⋅𝑡 𝑑𝑡
Integration by parts: 𝑢=𝑡, 𝑑𝑣=𝑒3𝑡 𝑑𝑡:
=𝑡𝑒3𝑡
3−𝑒3𝑡
9
Substitute back 𝑡=ln(𝑥):
=𝑥3(ln(𝑥)−1
3)
3+𝐶
Problem 2
Evaluate the integral:
∫𝑥𝑒2𝑥 𝑑𝑥
Solution
Using integration by parts, let 𝑢=𝑥 and 𝑑𝑣 =𝑒2𝑥 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=𝑒2𝑥
2
∫𝑥𝑒2𝑥 𝑑𝑥=𝑥⋅𝑒2𝑥
2−∫ 𝑒2𝑥
2 𝑑𝑥
=𝑥𝑒2𝑥
2−1
2⋅𝑒2𝑥
2
=𝑥𝑒2𝑥
2−𝑒2𝑥
4
=𝑒2𝑥(2𝑥−1)
4+𝐶
Verification
Differentiate 𝐹(𝑥)=𝑒2𝑥(2𝑥−1)
4:
𝐹′(𝑥)=1
4(𝑒2𝑥(2𝑥−1)⋅2+𝑒2𝑥 ⋅2)
=𝑥𝑒2𝑥
Alternative Approach
Substitute 𝑢=2𝑥, 𝑑𝑢=2𝑑𝑥, 𝑑𝑥=𝑑𝑢
2:
∫𝑥𝑒2𝑥 𝑑𝑥=1
2∫𝑢
2𝑒𝑢 𝑑𝑢
Integration by parts: 𝑣=𝑢
2, 𝑑𝑤=𝑒𝑢:
=1
2(𝑢𝑒𝑢
2−∫ 𝑒𝑢
2 𝑑𝑢)
Substitute back 𝑢=2𝑥:
=𝑒2𝑥(2𝑥−1)
4+𝐶
Problem 3
Evaluate the integral:
∫sin2(𝑥) 𝑑𝑥
Solution
Using the identity sin2(𝑥)=1−cos(2𝑥)
2:
∫sin2(𝑥) 𝑑𝑥=∫ 1−cos(2𝑥)
2 𝑑𝑥
=1
2(∫1 𝑑𝑥−∫cos(2𝑥) 𝑑𝑥)
=1
2(𝑥−sin(2𝑥)
2)+𝐶
=𝑥
2−sin(2𝑥)
4+𝐶
Verification
Differentiate 𝐹(𝑥)=𝑥
2−sin(2𝑥)
4:
𝐹′(𝑥)=1
2−1
4⋅2cos(2𝑥)
=1
2−1
2cos(2𝑥)
=sin2(𝑥)
Alternative Approach
Use the double-angle identity:
∫sin2(𝑥) 𝑑𝑥=∫ 1−cos(2𝑥)
2 𝑑𝑥
=1
2(𝑥−sin(2𝑥)
2)+𝐶
Problem 4
Evaluate the integral:
∫𝑥3𝑒𝑥 𝑑𝑥
Solution
Using integration by parts, let 𝑢=𝑥3 and 𝑑𝑣=𝑒𝑥 𝑑𝑥:
𝑑𝑢=3𝑥2 𝑑𝑥 and 𝑣=𝑒𝑥
∫𝑥3𝑒𝑥 𝑑𝑥=𝑥3𝑒𝑥−∫3𝑥2𝑒𝑥 𝑑𝑥
Apply integration by parts again:
=𝑥3𝑒𝑥−3(𝑥2𝑒𝑥−∫2𝑥𝑒𝑥 𝑑𝑥)
=𝑥3𝑒𝑥−3𝑥2𝑒𝑥+6(𝑥𝑒𝑥−∫𝑒𝑥 𝑑𝑥)
=𝑥3𝑒𝑥−3𝑥2𝑒𝑥+6𝑥𝑒𝑥−6𝑒𝑥+𝐶
=𝑒𝑥(𝑥3−3𝑥2+6𝑥−6)+𝐶
Verification
Differentiate 𝐹(𝑥)=𝑒𝑥(𝑥3−3𝑥2+6𝑥−6):
𝐹′(𝑥)=𝑒𝑥(𝑥3−3𝑥2+6𝑥−6)+𝑒𝑥(3𝑥2−6𝑥+6)
=𝑥3𝑒𝑥
Alternative Approach
Using substitution, 𝑢=𝑥, 𝑑𝑣 =𝑥2𝑒𝑥 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫𝑥2𝑒𝑥 𝑑𝑥
Evaluate ∫𝑥2𝑒𝑥 𝑑𝑥 as in Problem 1:
=𝑒𝑥(𝑥2−2𝑥+2)
Substitute back:
=𝑥3𝑒𝑥−3𝑥2𝑒𝑥+6𝑥𝑒𝑥−6𝑒𝑥+𝐶
Problem 5
Evaluate the integral:
∫ln(𝑥) 𝑑𝑥
Solution
Using integration by parts, let 𝑢=ln(𝑥) and 𝑑𝑣=𝑑𝑥:
𝑑𝑢=1
𝑥 𝑑𝑥 and 𝑣=𝑥
∫ln(𝑥) 𝑑𝑥=𝑥ln(𝑥)−∫𝑥⋅1
𝑥 𝑑𝑥
=𝑥ln(𝑥)−∫1 𝑑𝑥
=𝑥ln(𝑥)−𝑥+𝐶
Verification
Differentiate 𝐹(𝑥)=𝑥ln(𝑥)−𝑥:
𝐹′(𝑥)=ln(𝑥)+𝑥⋅1
𝑥−1
=ln(𝑥)
Alternative Approach
Using substitution, 𝑥 =𝑒𝑡, 𝑑𝑥=𝑒𝑡 𝑑𝑡:
∫ln(𝑥) 𝑑𝑥=∫𝑡𝑒𝑡 𝑑𝑡
Integration by parts: 𝑢=𝑡, 𝑑𝑣=𝑒𝑡 𝑑𝑡:
=𝑡𝑒𝑡−∫𝑒𝑡 𝑑𝑡
Substitute back 𝑡=ln(𝑥):
=𝑥ln(𝑥)−𝑥+𝐶
Problem 6
Evaluate the integral:
∫𝑒𝑥sin(𝑥) 𝑑𝑥
Solution
Using integration by parts, let 𝑢=𝑒𝑥 and 𝑑𝑣=sin(𝑥) 𝑑𝑥:
𝑑𝑢=𝑒𝑥 𝑑𝑥 and 𝑣=−cos(𝑥)
∫𝑒𝑥sin(𝑥) 𝑑𝑥=−𝑒𝑥cos(𝑥)−∫ −𝑒𝑥cos(𝑥) 𝑑𝑥
Applying integration by parts again:
=−𝑒𝑥cos(𝑥)+∫𝑒𝑥cos(𝑥) 𝑑𝑥
=−𝑒𝑥cos(𝑥)+𝑒𝑥sin(𝑥)−∫𝑒𝑥sin(𝑥) 𝑑𝑥
Setting up the equation:
𝐼=∫𝑒𝑥sin(𝑥) 𝑑𝑥
𝐼=−𝑒𝑥cos(𝑥)+𝑒𝑥sin(𝑥)−𝐼
2𝐼=𝑒𝑥(sin(𝑥)−cos(𝑥))
𝐼=𝑒𝑥(sin(𝑥)−cos(𝑥))
2+𝐶
Verification
Differentiate 𝐹(𝑥)=𝑒𝑥(sin(𝑥)−cos(𝑥))
2:
𝐹′(𝑥)=1
2(𝑒𝑥(sin(𝑥)−cos(𝑥))+𝑒𝑥(cos(𝑥)+sin(𝑥)))
=𝑒𝑥sin(𝑥)
Alternative Approach
Use Euler’s formula, 𝑒𝑖𝑥 =cos(𝑥)+𝑖sin(𝑥):
∫𝑒𝑥sin(𝑥) 𝑑𝑥=Im(∫𝑒(1+𝑖)𝑥 𝑑𝑥)
=Im (𝑒(1+𝑖)𝑥
1+𝑖 )
Simplify and separate real and imaginary parts:
=𝑒𝑥(sin(𝑥)−cos(𝑥))
2+𝐶
Problem 7
Evaluate the integral:
∫1
𝑥2+𝑎2 𝑑𝑥
Solution
Using the substitution 𝑥=𝑎tan(𝜃):
𝑑𝑥=𝑎sec2(𝜃) 𝑑𝜃
∫1
𝑥2+𝑎2 𝑑𝑥=∫ 1
𝑎2tan2(𝜃)+𝑎2⋅𝑎sec2(𝜃) 𝑑𝜃
=1
𝑎∫sec2(𝜃)
sec2(𝜃) 𝑑𝜃
=1
𝑎∫𝑑𝜃
=𝜃
𝑎+𝐶
Substitute back 𝜃=tan−1(𝑥
𝑎):
=1
𝑎tan−1(𝑥
𝑎)+𝐶
Verification
Differentiate 𝐹(𝑥)=1
𝑎tan−1(𝑥
𝑎):
𝐹′(𝑥)=1
𝑎⋅1
1+(𝑥
𝑎)2⋅1
𝑎
=1
𝑥2+𝑎2
Alternative Approach
Using partial fractions, 𝑥=𝑎:
∫1
𝑥2+𝑎2 𝑑𝑥=∫ 1/𝑎2
1+(𝑥
𝑎)2 𝑑𝑥
Substitute 𝑢=𝑥
𝑎, 𝑑𝑢=𝑑𝑥
𝑎:
=1
𝑎∫1
1+𝑢2 𝑑𝑢
=1
𝑎tan−1(𝑢)+𝐶
Substitute back 𝑢=𝑥
𝑎:
=1
𝑎tan−1(𝑥
𝑎)+𝐶
Problem 8
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Alternative Approach
Using integration by parts, let 𝑢=𝑥 and 𝑑𝑣 =𝑒−𝑥2 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫𝑒−𝑥2 𝑑𝑥
Apply integration by parts:
=𝑥∫𝑒−𝑥2 𝑑𝑥−∫1⋅∫𝑒−𝑥2 𝑑𝑥 𝑑𝑥
=−𝑒−𝑥2
2+𝐶
Problem 10
Evaluate the integral:
∫𝑒𝑥
1+𝑒𝑥 𝑑𝑥
Solution
Using the substitution 𝑢=1+𝑒𝑥:
𝑑𝑢=𝑒𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
𝑒𝑥
∫𝑒𝑥
1+𝑒𝑥 𝑑𝑥=∫ 1
𝑢 𝑑𝑢
=ln|𝑢|+𝐶
Substitute back 𝑢=1+𝑒𝑥:
=ln|1+𝑒𝑥|+𝐶
Verification
Differentiate 𝐹(𝑥)=ln|1+𝑒𝑥|:
𝐹′(𝑥)=1
1+𝑒𝑥⋅𝑒𝑥
=𝑒𝑥
1+𝑒𝑥
Alternative Approach
Using partial fractions, 𝑥=𝑒𝑥:
∫𝑒𝑥
1+𝑒𝑥 𝑑𝑥=∫ 1
1+𝑒𝑥⋅𝑒𝑥 𝑑𝑥
Substitute 𝑢=1+𝑒𝑥:
=ln|1+𝑒𝑥|+𝐶
Integration Problems with Detailed Solutions, Verification, and
Alternative Methods
Problem 1
Evaluate the integral:
∫𝑥2ln(𝑥) 𝑑𝑥
Solution
Using integration by parts, let 𝑢=ln(𝑥) and 𝑑𝑣=𝑥2 𝑑𝑥:
𝑑𝑢=1
𝑥 𝑑𝑥 and 𝑣=𝑥3
3
∫𝑥2ln(𝑥) 𝑑𝑥=ln(𝑥)⋅𝑥3
3−∫ 𝑥3
3⋅1
𝑥 𝑑𝑥
=𝑥3ln(𝑥)
3−1
3∫𝑥2 𝑑𝑥
=𝑥3ln(𝑥)
3−1
3⋅𝑥3
3
=𝑥3ln(𝑥)
3−𝑥3
9
=𝑥3(ln(𝑥)−1
3)
3+𝐶
Verification
Differentiate 𝐹(𝑥)=𝑥3(ln(𝑥)−1
3)
3:
𝐹′(𝑥)=1
3(3𝑥2(ln(𝑥)−1
3)+𝑥3⋅1
𝑥)
=𝑥2ln(𝑥)
Alternative Approach
Substitute 𝑥 =𝑒𝑡, 𝑑𝑥=𝑒𝑡 𝑑𝑡:
∫𝑥2ln(𝑥) 𝑑𝑥=∫𝑒3𝑡 ⋅𝑡 𝑑𝑡
Integration by parts: 𝑢=𝑡, 𝑑𝑣=𝑒3𝑡 𝑑𝑡:
=𝑡𝑒3𝑡
3−𝑒3𝑡
9
Substitute back 𝑡=ln(𝑥):
=𝑥3(ln(𝑥)−1
3)
3+𝐶
Problem 2
Evaluate the integral:
∫𝑥𝑒2𝑥 𝑑𝑥
Solution
Using integration by parts, let 𝑢=𝑥 and 𝑑𝑣 =𝑒2𝑥 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=𝑒2𝑥
2
∫𝑥𝑒2𝑥 𝑑𝑥=𝑥⋅𝑒2𝑥
2−∫ 𝑒2𝑥
2 𝑑𝑥
=𝑥𝑒2𝑥
2−1
2⋅𝑒2𝑥
2
=𝑥𝑒2𝑥
2−𝑒2𝑥
4
=𝑒2𝑥(2𝑥−1)
4+𝐶
Verification
Differentiate 𝐹(𝑥)=𝑒2𝑥(2𝑥−1)
4:
𝐹′(𝑥)=1
4(𝑒2𝑥(2𝑥−1)⋅2+𝑒2𝑥 ⋅2)
=𝑥𝑒2𝑥
Alternative Approach
Substitute 𝑢=2𝑥, 𝑑𝑢=2𝑑𝑥, 𝑑𝑥=𝑑𝑢
2:
∫𝑥𝑒2𝑥 𝑑𝑥=1
2∫𝑢
2𝑒𝑢 𝑑𝑢
Integration by parts: 𝑣=𝑢
2, 𝑑𝑤=𝑒𝑢:
=1
2(𝑢𝑒𝑢
2−∫ 𝑒𝑢
2 𝑑𝑢)
Substitute back 𝑢=2𝑥:
=𝑒2𝑥(2𝑥−1)
4+𝐶
Problem 3
Evaluate the integral:
∫sin2(𝑥) 𝑑𝑥
Solution
Using the identity sin2(𝑥)=1−cos(2𝑥)
2:
∫sin2(𝑥) 𝑑𝑥=∫ 1−cos(2𝑥)
2 𝑑𝑥
=1
2(∫1 𝑑𝑥−∫cos(2𝑥) 𝑑𝑥)
=1
2(𝑥−sin(2𝑥)
2)+𝐶
=𝑥
2−sin(2𝑥)
4+𝐶
Verification
Differentiate 𝐹(𝑥)=𝑥
2−sin(2𝑥)
4:
𝐹′(𝑥)=1
2−1
4⋅2cos(2𝑥)
=1
2−1
2cos(2𝑥)
=sin2(𝑥)
Alternative Approach
Use the double-angle identity:
∫sin2(𝑥) 𝑑𝑥=∫ 1−cos(2𝑥)
2 𝑑𝑥
=1
2(𝑥−sin(2𝑥)
2)+𝐶
Problem 4
Evaluate the integral:
∫𝑥3𝑒𝑥 𝑑𝑥
Solution
Using integration by parts, let 𝑢=𝑥3 and 𝑑𝑣=𝑒𝑥 𝑑𝑥:
𝑑𝑢=3𝑥2 𝑑𝑥 and 𝑣=𝑒𝑥
∫𝑥3𝑒𝑥 𝑑𝑥=𝑥3𝑒𝑥−∫3𝑥2𝑒𝑥 𝑑𝑥
Apply integration by parts again:
=𝑥3𝑒𝑥−3(𝑥2𝑒𝑥−∫2𝑥𝑒𝑥 𝑑𝑥)
=𝑥3𝑒𝑥−3𝑥2𝑒𝑥+6(𝑥𝑒𝑥−∫𝑒𝑥 𝑑𝑥)
=𝑥3𝑒𝑥−3𝑥2𝑒𝑥+6𝑥𝑒𝑥−6𝑒𝑥+𝐶
=𝑒𝑥(𝑥3−3𝑥2+6𝑥−6)+𝐶
Verification
Differentiate 𝐹(𝑥)=𝑒𝑥(𝑥3−3𝑥2+6𝑥−6):
𝐹′(𝑥)=𝑒𝑥(𝑥3−3𝑥2+6𝑥−6)+𝑒𝑥(3𝑥2−6𝑥+6)
=𝑥3𝑒𝑥
Alternative Approach
Using substitution, 𝑢=𝑥, 𝑑𝑣 =𝑥2𝑒𝑥 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫𝑥2𝑒𝑥 𝑑𝑥
Evaluate ∫𝑥2𝑒𝑥 𝑑𝑥 as in Problem 1:
=𝑒𝑥(𝑥2−2𝑥+2)
Substitute back:
=𝑥3𝑒𝑥−3𝑥2𝑒𝑥+6𝑥𝑒𝑥−6𝑒𝑥+𝐶
Problem 5
Evaluate the integral:
∫ln(𝑥) 𝑑𝑥
Solution
Using integration by parts, let 𝑢=ln(𝑥) and 𝑑𝑣=𝑑𝑥:
𝑑𝑢=1
𝑥 𝑑𝑥 and 𝑣=𝑥
∫ln(𝑥) 𝑑𝑥=𝑥ln(𝑥)−∫𝑥⋅1
𝑥 𝑑𝑥
=𝑥ln(𝑥)−∫1 𝑑𝑥
=𝑥ln(𝑥)−𝑥+𝐶
Verification
Differentiate 𝐹(𝑥)=𝑥ln(𝑥)−𝑥:
𝐹′(𝑥)=ln(𝑥)+𝑥⋅1
𝑥−1
=ln(𝑥)
Alternative Approach
Using substitution, 𝑥 =𝑒𝑡, 𝑑𝑥=𝑒𝑡 𝑑𝑡:
∫ln(𝑥) 𝑑𝑥=∫𝑡𝑒𝑡 𝑑𝑡
Integration by parts: 𝑢=𝑡, 𝑑𝑣=𝑒𝑡 𝑑𝑡:
=𝑡𝑒𝑡−∫𝑒𝑡 𝑑𝑡
Substitute back 𝑡=ln(𝑥):
=𝑥ln(𝑥)−𝑥+𝐶
Problem 6
Evaluate the integral:
∫𝑒𝑥sin(𝑥) 𝑑𝑥
Solution
Using integration by parts, let 𝑢=𝑒𝑥 and 𝑑𝑣=sin(𝑥) 𝑑𝑥:
𝑑𝑢=𝑒𝑥 𝑑𝑥 and 𝑣=−cos(𝑥)
∫𝑒𝑥sin(𝑥) 𝑑𝑥=−𝑒𝑥cos(𝑥)−∫ −𝑒𝑥cos(𝑥) 𝑑𝑥
Applying integration by parts again:
=−𝑒𝑥cos(𝑥)+∫𝑒𝑥cos(𝑥) 𝑑𝑥
=−𝑒𝑥cos(𝑥)+𝑒𝑥sin(𝑥)−∫𝑒𝑥sin(𝑥) 𝑑𝑥
Setting up the equation:
𝐼=∫𝑒𝑥sin(𝑥) 𝑑𝑥
𝐼=−𝑒𝑥cos(𝑥)+𝑒𝑥sin(𝑥)−𝐼
2𝐼=𝑒𝑥(sin(𝑥)−cos(𝑥))
𝐼=𝑒𝑥(sin(𝑥)−cos(𝑥))
2+𝐶
Verification
Differentiate 𝐹(𝑥)=𝑒𝑥(sin(𝑥)−cos(𝑥))
2:
𝐹′(𝑥)=1
2(𝑒𝑥(sin(𝑥)−cos(𝑥))+𝑒𝑥(cos(𝑥)+sin(𝑥)))
=𝑒𝑥sin(𝑥)
Alternative Approach
Use Euler’s formula, 𝑒𝑖𝑥 =cos(𝑥)+𝑖sin(𝑥):
∫𝑒𝑥sin(𝑥) 𝑑𝑥=Im(∫𝑒(1+𝑖)𝑥 𝑑𝑥)
=Im (𝑒(1+𝑖)𝑥
1+𝑖 )
Simplify and separate real and imaginary parts:
=𝑒𝑥(sin(𝑥)−cos(𝑥))
2+𝐶
Problem 7
Evaluate the integral:
∫1
𝑥2+𝑎2 𝑑𝑥
Solution
Using the substitution 𝑥=𝑎tan(𝜃):
𝑑𝑥=𝑎sec2(𝜃) 𝑑𝜃
∫1
𝑥2+𝑎2 𝑑𝑥=∫ 1
𝑎2tan2(𝜃)+𝑎2⋅𝑎sec2(𝜃) 𝑑𝜃
=1
𝑎∫sec2(𝜃)
sec2(𝜃) 𝑑𝜃
=1
𝑎∫𝑑𝜃
=𝜃
𝑎+𝐶
Substitute back 𝜃=tan−1(𝑥
𝑎):
=1
𝑎tan−1(𝑥
𝑎)+𝐶
Verification
Differentiate 𝐹(𝑥)=1
𝑎tan−1(𝑥
𝑎):
𝐹′(𝑥)=1
𝑎⋅1
1+(𝑥
𝑎)2⋅1
𝑎
=1
𝑥2+𝑎2
Alternative Approach
Using partial fractions, 𝑥=𝑎:
∫1
𝑥2+𝑎2 𝑑𝑥=∫ 1/𝑎2
1+(𝑥
𝑎)2 𝑑𝑥
Substitute 𝑢=𝑥
𝑎, 𝑑𝑢=𝑑𝑥
𝑎:
=1
𝑎∫1
1+𝑢2 𝑑𝑢
=1
𝑎tan−1(𝑢)+𝐶
Substitute back 𝑢=𝑥
𝑎:
=1
𝑎tan−1(𝑥
𝑎)+𝐶
Problem 8
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Evaluate the integral:
∫𝑥cos(𝑥2) 𝑑𝑥
Solution
Using the substitution 𝑢=𝑥2:
𝑑𝑢=2𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
2𝑥
∫𝑥cos(𝑥2) 𝑑𝑥=1
2∫cos(𝑢) 𝑑𝑢
=1
2sin(𝑢)+𝐶
Substitute back 𝑢=𝑥2:
=sin(𝑥2)
2+𝐶
Verification
Differentiate 𝐹(𝑥)=sin(𝑥2)
2:
𝐹′(𝑥)=1
2⋅cos(𝑥2)⋅2𝑥
=𝑥cos(𝑥2)
Alternative Approach
Using another substitution, 𝑢=𝑥, 𝑑𝑣=cos(𝑥2) 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫cos(𝑥2) 𝑑𝑥
Apply integration by parts:
=𝑥∫cos(𝑥2) 𝑑𝑥−∫1⋅∫cos(𝑥2) 𝑑𝑥 𝑑𝑥
=𝑥sin(𝑥2)−∫sin(𝑥2) 𝑑𝑥
=sin(𝑥2)
2+𝐶
Problem 9
Evaluate the integral:
∫𝑥𝑒−𝑥2 𝑑𝑥
Solution
Using the substitution 𝑢=−𝑥2:
𝑑𝑢=−2𝑥 𝑑𝑥⟹𝑑𝑥 = 𝑑𝑢
−2𝑥
∫𝑥𝑒−𝑥2 𝑑𝑥=−1
2∫𝑒𝑢 𝑑𝑢
=−1
2𝑒𝑢+𝐶
Substitute back 𝑢=−𝑥2:
=−𝑒−𝑥2
2+𝐶
Verification
Differentiate 𝐹(𝑥)=−𝑒−𝑥2
2:
𝐹′(𝑥)=−1
2⋅−2𝑥𝑒−𝑥2
=𝑥𝑒−𝑥2
Alternative Approach
Using integration by parts, let 𝑢=𝑥 and 𝑑𝑣 =𝑒−𝑥2 𝑑𝑥:
𝑑𝑢=𝑑𝑥 and 𝑣=∫𝑒−𝑥2 𝑑𝑥
Apply integration by parts:
=𝑥∫𝑒−𝑥2 𝑑𝑥−∫1⋅∫𝑒−𝑥2 𝑑𝑥 𝑑𝑥
=−𝑒−𝑥2
2+𝐶
Problem 10
Evaluate the integral:
∫𝑒𝑥
1+𝑒𝑥 𝑑𝑥
Solution
Using the substitution 𝑢=1+𝑒𝑥:
𝑑𝑢=𝑒𝑥 𝑑𝑥⟹𝑑𝑥 =𝑑𝑢
𝑒𝑥
∫𝑒𝑥
1+𝑒𝑥 𝑑𝑥=∫ 1
𝑢 𝑑𝑢
=ln|𝑢|+𝐶
Substitute back 𝑢=1+𝑒𝑥:
=ln|1+𝑒𝑥|+𝐶
Verification
Differentiate 𝐹(𝑥)=ln|1+𝑒𝑥|:
𝐹′(𝑥)=1
1+𝑒𝑥⋅𝑒𝑥
=𝑒𝑥
1+𝑒𝑥
Alternative Approach
Using partial fractions, 𝑥=𝑒𝑥:
∫𝑒𝑥
1+𝑒𝑥 𝑑𝑥=∫ 1
1+𝑒𝑥⋅𝑒𝑥 𝑑𝑥
Substitute 𝑢=1+𝑒𝑥:
=ln|1+𝑒𝑥|+𝐶
