Module 8 Practice Problems: Techniques of Integration
Instructions: Work through each problem on paper first, then reveal the solution to check your work.
Problem 1
Evaluate ∫ cos(5x + 1) dx using u-substitution.
Solution
u = 5x + 1, du = 5 dx. (1/5) ∫ cos u du = (1/5) sin(5x + 1) + C.
Problem 2
Evaluate ∫01 x(1 − x²)3 dx.
Solution
u = 1 − x², du = −2x dx. Bounds: u(0) = 1, u(1) = 0. (−1/2) ∫10 u³ du = (1/2) ∫01 u³ du = (1/2)(1/4) = 1/8.
Problem 3
Evaluate ∫ sin4x cos x dx.
Solution
u = sin x, du = cos x dx. ∫ u4 du = u5/5 + C = sin5x / 5 + C.
Problem 4
Evaluate ∫ sin²(3x) dx.
Solution
Power reduction: sin²(3x) = (1 − cos 6x)/2. ∫ (1 − cos 6x)/2 dx = x/2 − sin(6x)/12 + C.
Problem 5
Evaluate ∫ x² ln x dx using integration by parts.
Solution
u = ln x (L), dv = x² dx. du = (1/x) dx, v = x³/3. = (x³/3) ln x − ∫ x²/3 dx = (x³/3) ln x − x³/9 + C = (x³/9)(3 ln x − 1) + C.
Problem 6
Evaluate ∫ x e2x dx.
Solution
u = x, dv = e2x dx. du = dx, v = e2x/2. = xe2x/2 − ∫ e2x/2 dx = xe2x/2 − e2x/4 + C = (e2x/4)(2x − 1) + C.
Problem 7
Evaluate ∫ ex cos x dx.
Solution
Boomerang: IBP twice. I = ex cos x + ex sin x − I. So 2I = ex(cos x + sin x). I = (ex/2)(cos x + sin x) + C.
Problem 8
Use the Trapezoidal Rule with n = 4 to approximate ∫02 √(1 + x³) dx.
Solution
Δx = 0.5. Values: f(0) = 1, f(0.5) = √1.125 ≈ 1.0607, f(1) = √2 ≈ 1.4142, f(1.5) = √4.375 ≈ 2.0917, f(2) = 3. T4 = 0.25[1 + 2(1.0607) + 2(1.4142) + 2(2.0917) + 3] = 0.25(13.133) ≈ 3.283.
Problem 9
Use Simpson's Rule with n = 4 for ∫02 √(1 + x³) dx (same as Problem 8).
Solution
S4 = (0.5/3)[1 + 4(1.0607) + 2(1.4142) + 4(2.0917) + 3] = (1/6)[1 + 4.2428 + 2.8284 + 8.3668 + 3] = (1/6)(19.438) ≈ 3.240.
Problem 10
Evaluate ∫0π/2 x cos x dx using IBP.
Solution
u = x, dv = cos x dx. [x sin x + cos x]0π/2 = (π/2 + 0) − (0 + 1) = π/2 − 1.