Advanced Integration by Substitution: Composites, Roots & Trig Powers

Notice that the integrand is 3x² cos(x³). The derivative of the inside function x³ is 3x², which is exactly the factor in front of the cosine. This is the reverse chain rule pattern. Let u = x³, then du/dx = 3x² so du = 3x² dx. The integral becomes ∫ cos(u) du. The antiderivative of cos(u) is sin(u), so ∫ cos(u) du = sin(u) + C. Substituting back u = x³ gives sin(x³) + C.

The integrand is 4x (x² + 1)⁵. The derivative of the inside x² + 1 is 2x, so we are close to the chain rule in reverse. Let u = x² + 1, then du/dx = 2x so du = 2x dx, which means 2x dx = du. Rewrite 4x dx as 2 · 2x dx = 2 du. The integral becomes ∫ 2 u⁵ du = 2 · (u⁶ / 6) + C = (⅓) u⁶ + C. Substituting back u = x² + 1 gives (⅓)(x² + 1)⁶ + C.

Let u = 4 - x², then du/dx = -2x so du = -2x dx, which means x dx = -1/2 du. Change the limits: when x = 0, u = 4; when x = 2, u = 0. The integral becomes ∫ from u=4 to u=0 of √u · (-½) du = (½) ∫₀⁴ √u du = (½) ∫₀⁴ u½ du = (½) · (⅔) u³/2 evaluated from 0 to 4 = (⅓) [4^(3/2) - 0] = (⅓)(8) = 8/3.

tan(x) = sin(x)/cos(x). Let u = cos(x), then du = -sin(x) dx, so sin(x) dx = -du. The integral becomes ∫ (sin(x)/cos(x)) dx = ∫ (1/u) (-du) = -∫ (1/u) du = -ln|u| + C = -ln|cos(x)| + C. Using properties of logarithms, -ln|cos(x)| = ln|1/cos(x)| = ln|sec(x)| + C.

The integrand is e^(3x). The derivative of 3x is 3, so we need to compensate. Let u = 3x, then du/dx = 3 so du = 3 dx, which means dx = du/3. The integral becomes ∫ e^u · (du/3) = (⅓) ∫ e^u du = (⅓) e^u + C. Substituting back u = 3x gives (⅓) e^(3x) + C.

Rewrite the integrand as x^(-½). Let u = x, then du = dx. Limits: when x = 1, u = 1; when x = 4, u = 4. The integral is ∫ from 1 to 4 of u^(-½) du = [2 u^(½)] from 1 to 4 = 2√4 - 2√1 = 2·2 - 2·1 = 4 - 2 = 2. Alternatively, let u = √x, then u² = x, 2u du = dx. When x = 1, u = 1; when x = 4, u = 2. The integral becomes ∫ from 1 to 2 of (1/u) · 2u du = ∫ from 1 to 2 of 2 du = 2[u] from 1 to 2 = 2(2 - 1) = 2.

The numerator x is almost the derivative of the denominator x² + 5 (whose derivative is 2x). Let u = x² + 5, then du = 2x dx, so x dx = (½) du. The integral becomes ∫ (1/u) · (½) du = (½) ∫ (1/u) du = (½) ln|u| + C = (½) ln|x² + 5| + C.

Let u = sin(2x), then du = 2 cos(2x) dx, so cos(2x) dx = du/2. The integral becomes ∫ u · (du/2) = (½) ∫ u du = (½)(u²/2) + C = (1/4)u² + C = (1/4) sin²(2x) + C.

The numerator x³ is almost the derivative of x⁴ + 1 (which is 4x³). Let u = x⁴ + 1, then du = 4x³ dx, so x³ dx = du/4. The integral becomes ∫ (1/√u) · (du/4) = (1/4) ∫ u^(-½) du = (1/4) · 2 u^(½) + C = (½) √u + C = (½) √(x⁴ + 1) + C.

Let u = ln x, then du = (1/x) dx, so dx/x = du. The integral becomes ∫ du / √u = ∫ u^(-½) du = 2 u^(½) + C = 2 √(ln x) + C.

Let u = x³ + 2, then du = 3x² dx, so x² dx = du/3. The integral becomes ∫ u⁴ · (du/3) = (⅓) ∫ u⁴ du = (⅓) (u⁵ / 5) + C = u⁵ / 15 + C = (x³ + 2)⁵ / 15 + C.

sin³(x) = sin(x) (1 - cos²(x)). Let u = cos(x), then du = -sin(x) dx, so sin(x) dx = -du. The integral becomes ∫ (1 - u²) (-du) = -∫ (1 - u²) du = - (u - u³/3) + C = -u + (⅓) u³ + C = -cos(x) + (⅓) cos³(x) + C.

We can solve this using substitution. Let u = x² - 1. Then differentiate to find du/dx = 2x, which means du = 2x dx. Our integrand contains exactly 2x dx, so we can substitute directly. The integral becomes ∫u³ du = u⁴/4 + C. Now we substitute back u = x² - 1 to obtain (x² - 1)⁴/4 + C.

We use substitution to evaluate this definite integral. Let u = sin(3x). Then differentiate to find du/dx = 3cos(3x), which means du = 3cos(3x) dx, or equivalently cos(3x) dx = du/3. When x = 0, u = sin(0) = 0. When x = π/6, u = sin(π/2) = 1. Substituting into the integral with the new limits, we get ∫from u=0 to u=1 of u²(du/3) = (⅓)∫u² du from 0 to 1 = (⅓)[u³/3]from 0 to 1 = (⅓)(1/3 - 0) = 1/9.

The substitution u = x would not be effective because it doesn't simplify the expression. We would get ∫(u² + 1)^(½) du, which is the same as the original integral and doesn't make progress toward solving it. The other substitutions are all valid approaches: u = x² + 1 gives us an integral requiring further techniques, u = tan(θ) where x = tan(θ) leads to trigonometric substitution which simplifies to ∫sec³(θ) dθ, and u = sinh(θ) where x = sinh(θ) leads to hyperbolic substitution which simplifies to ∫cosh²(θ) dθ.