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

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.