Integration by Parts and Substitution Worksheet

Question 1

Evaluate ∫ 2x cos(x^2) dx using substitution.

Accepted Answer

Let u = x^2, so du = 2x dx. The integral becomes ∫ cos(u) du, which equals sin(u) + C. Substituting back gives sin(x^2) + C.

Question 2

Evaluate ∫ 3x^2/(x^3 + 5) dx using substitution.

Accepted Answer

Let u = x^3 + 5, so du = 3x^2 dx. The integral becomes ∫ 1/u du, which equals ln|u| + C. Substituting back gives ln|x^3 + 5| + C.

Question 3

Evaluate ∫ x e^x dx using integration by parts.

Accepted Answer

Use u = x and dv = e^x dx. Then du = dx and v = e^x. By integration by parts, ∫ x e^x dx = x e^x - ∫ e^x dx = x e^x - e^x + C. The answer is e^x(x - 1) + C.

Question 4

Evaluate ∫ ln(x) dx for x > 0 using integration by parts.

Accepted Answer

Use u = ln(x) and dv = dx. Then du = 1/x dx and v = x. By integration by parts, ∫ ln(x) dx = x ln(x) - ∫ 1 dx = x ln(x) - x + C.

Question 5

Evaluate the definite integral ∫ from 0 to 1 of 4x(2x^2 + 1)^3 dx.

Accepted Answer

Let u = 2x^2 + 1, so du = 4x dx. When x = 0, u = 1. When x = 1, u = 3. The integral becomes ∫ from 1 to 3 of u^3 du = u^4/4 from 1 to 3 = (81 - 1)/4 = 20.

Question 6

Evaluate ∫ x sin(3x) dx using integration by parts.

Accepted Answer

Use u = x and dv = sin(3x) dx. Then du = dx and v = -cos(3x)/3. The integral is -x cos(3x)/3 + ∫ cos(3x)/3 dx, which equals -x cos(3x)/3 + sin(3x)/9 + C.

Question 7

Choose the better method, substitution or integration by parts, and evaluate ∫ x/(x^2 + 9) dx.

Accepted Answer

Substitution is the better method because the derivative of x^2 + 9 is 2x. Let u = x^2 + 9, so du = 2x dx and x dx = du/2. The integral becomes 1/2 ∫ 1/u du = 1/2 ln|u| + C. The answer is 1/2 ln(x^2 + 9) + C.

Question 8

Evaluate ∫ e^(2x) sin(e^(2x)) dx using substitution.

Accepted Answer

Let u = e^(2x), so du = 2e^(2x) dx and e^(2x) dx = du/2. The integral becomes 1/2 ∫ sin(u) du = -1/2 cos(u) + C. Substituting back gives -1/2 cos(e^(2x)) + C.

Question 9

Evaluate the definite integral ∫ from 1 to e of ln(x) dx. Interpret the result as an area.

Accepted Answer

Using integration by parts, an antiderivative of ln(x) is x ln(x) - x. Evaluating from 1 to e gives (e ln(e) - e) - (1 ln(1) - 1) = (e - e) - (0 - 1) = 1. The area under y = ln(x) from x = 1 to x = e is 1 square unit.

Question 10

Evaluate ∫ x^2 e^x dx using integration by parts twice.

Accepted Answer

First use u = x^2 and dv = e^x dx. Then du = 2x dx and v = e^x, so the integral is x^2e^x - ∫ 2x e^x dx. For ∫ 2x e^x dx, use integration by parts to get 2(xe^x - e^x). Therefore, the original integral is x^2e^x - 2xe^x + 2e^x + C = e^x(x^2 - 2x + 2) + C.

Question 11

Evaluate ∫ cos(√x)/√x dx using substitution.

Accepted Answer

Let u = √x. Then du = 1/(2√x) dx, so dx/√x = 2 du. The integral becomes 2 ∫ cos(u) du = 2 sin(u) + C. Substituting back gives 2 sin(√x) + C.

Question 12

Evaluate ∫ x ln(x^2) dx for x > 0 using integration by parts.

Accepted Answer

Use u = ln(x^2) and dv = x dx. Then du = 2/x dx and v = x^2/2. By integration by parts, the integral is (x^2/2)ln(x^2) - ∫ (x^2/2)(2/x) dx. This simplifies to (x^2/2)ln(x^2) - ∫ x dx = (x^2/2)ln(x^2) - x^2/2 + C.

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