Integration By Substitution Fundamentals: Linear, Power & Chain-Rule Forms

1) Valuate the indefinite integral ∫(2x + 1)⁴ dx

(2x + 1)⁵/10 + C

(2x + 1)⁵/5 + C

2(2x + 1)⁵ + C

(2x + 1)³ + C

We can solve this using the reverse chain rule or substitution. Let u = 2x + 1. Then du/dx = 2, so du = 2dx, which means dx = du/2. Substituting, we get ∫u⁴(du/2) = (½)∫u⁴ du = (½)(u⁵/5) + C = u⁵/10 + C. However, we must substitute back u = 2x + 1 to get (2x + 1)⁵/10 + C.

Explanation

We can solve this using the reverse chain rule or substitution. Let u = 2x + 1. Then du/dx = 2, so du = 2dx, which means dx = du/2. Substituting, we get ∫u⁴(du/2) = (½)∫u⁴ du = (½)(u⁵/5) + C = u⁵/10 + C. However, we must substitute back u = 2x + 1 to get (2x + 1)⁵/10 + C.

2) Find ∫3sin(2x + 1)dx

-3/2cos(2x + 1) + C

-3cos(2x + 1) + C

3/2cos(2x + 1) + C

3cos(2x + 1) + C

Let u = 2x + 1. Then du/dx = 2, so du = 2dx, which means dx = du/2. Substituting, we get ∫3sin(u)(du/2) = (3/2)∫sin(u) du = (3/2)(-cos(u)) + C = -3/2cos(u) + C. Substituting back u = 2x + 1 gives us -3/2cos(2x + 1) + C.

Explanation

Let u = 2x + 1. Then du/dx = 2, so du = 2dx, which means dx = du/2. Substituting, we get ∫3sin(u)(du/2) = (3/2)∫sin(u) du = (3/2)(-cos(u)) + C = -3/2cos(u) + C. Substituting back u = 2x + 1 gives us -3/2cos(2x + 1) + C.

3) Compute ∫2x(x² + 3)⁶ dx

2(x² + 3)⁷/7 + C

(x² + 3)⁷/7 + C

2(x² + 3)⁶ + C

(x² + 3)⁷/14 + C

Let u = x² + 3. Then du/dx = 2x, so du = 2x dx. Substituting, we get ∫u⁶ du = u⁷/7 + C. Substituting back u = x² + 3 gives us (x² + 3)⁷/7 + C.

Explanation

Let u = x² + 3. Then du/dx = 2x, so du = 2x dx. Substituting, we get ∫u⁶ du = u⁷/7 + C. Substituting back u = x² + 3 gives us (x² + 3)⁷/7 + C.

4) Evaluate ∫cos(x)sin³(x)dx

Sin⁴(x)/4 + C

4sin⁴(x) + C

Sin⁴(x) + C

Sin³(x)/3 + C

Let u = sin(x). Then du/dx = cos(x), so du = cos(x) dx. Substituting, we get ∫u³ du = u⁴/4 + C. Substituting back u = sin(x) gives us sin⁴(x)/4 + C.

Explanation

Let u = sin(x). Then du/dx = cos(x), so du = cos(x) dx. Substituting, we get ∫u³ du = u⁴/4 + C. Substituting back u = sin(x) gives us sin⁴(x)/4 + C.

5) Find the antiderivative of f(x) = 4x(x² + 5)^-2

-2/(x² + 5) + C

-4/(x² + 5) + C

2/(x² + 5) + C

4/(x² + 5) + C

Let u = x² + 5. Then du/dx = 2x, so du = 2x dx, which means 4x dx = 2 du. Substituting, we get ∫u^-2 × 2 du = 2∫u^-2 du = 2(-u^-1) + C = -2/u + C. Substituting back u = x² + 5 gives us -2/(x² + 5) + C.

Explanation

Let u = x² + 5. Then du/dx = 2x, so du = 2x dx, which means 4x dx = 2 du. Substituting, we get ∫u^-2 × 2 du = 2∫u^-2 du = 2(-u^-1) + C = -2/u + C. Substituting back u = x² + 5 gives us -2/(x² + 5) + C.

6) Evaluate ∫x²√(x³ + 1)dx

(1/9)(x³ + 1)³ + C

(2/9)(x³ + 1)3/2 + C

(⅔)(x³ + 1)³ + C

(1/3)(x³ + 1)3/2 + C

Let u = x³+ 1. Then du/dx = 3x², so du = 3x² dx, which means x² dx = du/3. Substituting, we get ∫√u (du/3) = (⅓)∫u½ du = (⅓)(u3/2/(3/2)) + C = (⅓)(2u3/2/3) + C = (2/9)u3/2 + C. Substituting back u = x³+ 1 gives us (2/9)(x³+ 1)3/2 + C = 2(x³+ 1)3/2/9 + C.

Explanation

Let u = x³+ 1. Then du/dx = 3x², so du = 3x² dx, which means x² dx = du/3. Substituting, we get ∫√u (du/3) = (⅓)∫u½ du = (⅓)(u3/2/(3/2)) + C = (⅓)(2u3/2/3) + C = (2/9)u3/2 + C. Substituting back u = x³+ 1 gives us (2/9)(x³+ 1)3/2 + C = 2(x³+ 1)3/2/9 + C.

7) Compute ∫e3xdx

E3x/3 + C

3e3x + C

E3x + C

3e3x/2 + C

Let u = 3x. Then du/dx = 3, so du = 3dx, which means dx = du/3. Substituting, we get ∫eu(du/3) = (⅓)∫eu du = (⅓)eu + C. Substituting back u = 3x gives us e3x/3 + C.

Explanation

Let u = 3x. Then du/dx = 3, so du = 3dx, which means dx = du/3. Substituting, we get ∫eu(du/3) = (⅓)∫eu du = (⅓)eu + C. Substituting back u = 3x gives us e3x/3 + C.

8) Find ∫sec²(x)tan⁵(x)dx

Tan⁶(x)/6 + C

Tan⁶(x) + C

6tan⁶(x) + C

Tan⁵(x)/5 + C

Let u = tan(x). Then du/dx = sec²(x), so du = sec²(x) dx. Substituting, we get ∫u⁵ du = u⁶/6 + C. Substituting back u = tan(x) gives us tan⁶(x)/6 + C.

Explanation

Let u = tan(x). Then du/dx = sec²(x), so du = sec²(x) dx. Substituting, we get ∫u⁵ du = u⁶/6 + C. Substituting back u = tan(x) gives us tan⁶(x)/6 + C.

9) Which substitution would be most effective for evaluating ∫x√(1 - x²)dx?

U = 1 - x²

U = √(1 - x²)

U = x²

U = 1 - x

When we have x times a function of (1 - x²), the substitution u = 1 - x² simplifies the problem because the derivative of 1 - x² is -2x, which gives us the x in the integrand. Let u = 1 - x², then du = -2x dx, so x dx = -1/2 du. Substituting, we get ∫√u(-1/2 du) = -1/2 ∫u^1/2 du = -1/2(2u³/2/3) + C = -1/3u³/2 + C = -1/3(1 - x²)³/2 + C.

Explanation

When we have x times a function of (1 - x²), the substitution u = 1 - x² simplifies the problem because the derivative of 1 - x² is -2x, which gives us the x in the integrand. Let u = 1 - x², then du = -2x dx, so x dx = -1/2 du. Substituting, we get ∫√u(-1/2 du) = -1/2 ∫u^1/2 du = -1/2(2u³/2/3) + C = -1/3u³/2 + C = -1/3(1 - x²)³/2 + C.

10) What happens to the limits of integration when using substitution for definite integrals?

The limits remain the same

The limits change according to the substitution

The limits are always converted to positive values

The limits are divided by the constant factor in the substitution

When using substitution for definite integrals, we must change both the variable and the limits of integration. The new limits correspond to the old limits substituted into the substitution equation. For example, if we let u = g(x), then when x = a, u = g(a), and when x = b, u = g(b). This is necessary because we are changing the variable of integration from x to u, so we must use the corresponding values of u at the original x values.

Explanation

When using substitution for definite integrals, we must change both the variable and the limits of integration. The new limits correspond to the old limits substituted into the substitution equation. For example, if we let u = g(x), then when x = a, u = g(a), and when x = b, u = g(b). This is necessary because we are changing the variable of integration from x to u, so we must use the corresponding values of u at the original x values.

11) Evaluate ∫x/(x² + 4)dx

Ln|x² + 4|/2 + C

2ln|x² + 4| + C

Ln|x² + 4| + C

Ln|x| + C

Let u = x² + 4. Then du/dx = 2x, so du = 2x dx, which means x dx = du/2. Substituting, we get ∫(1/u)(du/2) = (½)∫du/u = (½)ln|u| + C. Substituting back u = x² + 4 gives us ln|x² + 4|/2 + C.

Explanation

Let u = x² + 4. Then du/dx = 2x, so du = 2x dx, which means x dx = du/2. Substituting, we get ∫(1/u)(du/2) = (½)∫du/u = (½)ln|u| + C. Substituting back u = x² + 4 gives us ln|x² + 4|/2 + C.

12) What type of substitution is best for ∫(x² + 1)^(3/2)x dx?

U = x² + 1

U = x

U = x³ + x

U = (x² + 1)^(3/2)

The integrand contains x times a function of (x² + 1), so letting u = x² + 1 makes sense because the derivative of u is 2x, which gives us the x in the integrand up to a constant factor. When we substitute, we get x dx = du/2, and the integrand becomes ∫u^(3/2)(du/2) = (½)∫u^(3/2) du, which is straightforward to integrate.

Explanation

The integrand contains x times a function of (x² + 1), so letting u = x² + 1 makes sense because the derivative of u is 2x, which gives us the x in the integrand up to a constant factor. When we substitute, we get x dx = du/2, and the integrand becomes ∫u^(3/2)(du/2) = (½)∫u^(3/2) du, which is straightforward to integrate.

13) Compute ∫(2x - 3)⁵ dx

(2x - 3)⁶/6 + C

2(2x - 3)⁶ + C

(2x - 3)⁶/12 + C

(2x - 3)⁶/4 + C

Let u = 2x - 3. Then du/dx = 2, so du = 2dx, which means dx = du/2. Substituting, we get ∫u⁵(du/2) = (½)∫u⁵ du = (½)(u⁶/6) + C = u⁶/12 + C. Substituting back u = 2x - 3 gives us (2x - 3)⁶/12 + C.

Explanation

Let u = 2x - 3. Then du/dx = 2, so du = 2dx, which means dx = du/2. Substituting, we get ∫u⁵(du/2) = (½)∫u⁵ du = (½)(u⁶/6) + C = u⁶/12 + C. Substituting back u = 2x - 3 gives us (2x - 3)⁶/12 + C.

14) Which expression represents the derivative of tan⁴(x) with respect to x?

4tan³(x)sec²(x)

4tan³(x)

4tan³(x)sec(x)

Tan²(x)sec²(x)

Using the chain rule, let u = tan(x), so we have u⁴. The derivative of u⁴ is 4u³, and the derivative of u = tan(x) is sec²(x). Therefore, by the chain rule, d/dx[tan⁴(x)] = 4tan³(x) × sec²(x). This understanding helps us recognize that ∫4tan³(x)sec²(x)dx = tan⁴(x) + C.

Explanation

Using the chain rule, let u = tan(x), so we have u⁴. The derivative of u⁴ is 4u³, and the derivative of u = tan(x) is sec²(x). Therefore, by the chain rule, d/dx[tan⁴(x)] = 4tan³(x) × sec²(x). This understanding helps us recognize that ∫4tan³(x)sec²(x)dx = tan⁴(x) + C.

15) Find the value of ∫ππ/2 of cos(x)sin²(x) dx

1/3

-1/3

1/6

-1/6

Let u = sin(x). Then du = cos(x) dx. When x = π/2, u = 1. When x = π, u = 0. Substituting, the integral becomes ∫01 u² du = [u³/3]10 = 0³/3 - 1³/3 = 0 - 1/3 = -1/3.

Explanation

Let u = sin(x). Then du = cos(x) dx. When x = π/2, u = 1. When x = π, u = 0. Substituting, the integral becomes ∫01 u² du = [u³/3]10 = 0³/3 - 1³/3 = 0 - 1/3 = -1/3.