Integration Techniques — Topic Review — Solutions

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This review covers all four lessons in the Integration Techniques topic: Integration by Substitution, Integration by Parts, Partial Fractions Integration, and Integration with Trig Identities. Questions span all difficulty levels and mirror the style of QCAA Specialist Maths examination items. Click each answer box to reveal the full worked solution.

Review Questions

1. Fluency

   Q1 — Substitution: power of a function

   Find ∫4x(x² + 5)³ dx using the substitution u = x² + 5.

   ▶ Show Answer

   Let u = x² + 5.   Then du = 2x dx, so 4x dx = 2 du.

   ∫4x(x² + 5)³ dx = 2∫u³ du = 2 × u&sup4;/4 + C = u&sup4;/2 + C

   = (x² + 5)&sup4; / 2 + C

2. Fluency

   Q2 — Substitution: trigonometric

   Find ∫sin³(x)cos(x) dx using u = sin x.

   ▶ Show Answer

   Let u = sin x.   Then du = cos x dx.

   ∫sin³(x)cos(x) dx = ∫u³ du = u&sup4;/4 + C

   = sin&sup4;(x) / 4 + C

3. Understanding

   Q3 — Substitution: logarithmic form

   Find ∫ (2x) / (x² + 3) dx.

   ▶ Show Answer

   Let u = x² + 3.   Then du = 2x dx.

   ∫ (2x)/(x² + 3) dx = ∫ (1/u) du = ln|u| + C

   = ln(x² + 3) + C   (x² + 3 > 0 always, so no absolute value needed)

4. Understanding

   Q4 — Substitution: definite integral with changed limits

   Evaluate ∫₁³ x²√(x³ + 1) dx. Change the limits of integration.

   ▶ Show Answer

   Let u = x³ + 1.   Then du = 3x² dx, so x² dx = du/3.

   Change limits: x = 1 → u = 1 + 1 = 2;   x = 3 → u = 27 + 1 = 28.

   ∫₂²⁸ √u × du/3 = (1/3) ∫₂²⁸ u^1/2 du

   = (1/3) × [2u^3/2/3]₂²⁸ = (2/9)[28^3/2 − 2^3/2]

   28^3/2 = 28√28 = 28 × 2√7 = 56√7;   2^3/2 = 2√2

   = (2/9)(56√7 − 2√2) ≈ (2/9)(148.22 − 2.83) ≈ 32.31

5. Fluency

   Q5 — Integration by Parts: polynomial × exponential

   Find ∫xe^x dx using integration by parts.

   ▶ Show Answer

   Let u = x, dv = e^x dx.   Then du = dx, v = e^x.

   IBP formula: ∫u dv = uv − ∫v du

   ∫xe^x dx = xe^x − ∫e^x dx = xe^x − e^x + C

   = e^x(x − 1) + C

6. Understanding

   Q6 — Integration by Parts: polynomial × logarithm

   Find ∫x ln x dx.

   ▶ Show Answer

   Let u = ln x, dv = x dx.   Then du = (1/x) dx, v = x²/2.

   ∫x ln x dx = (x²/2) ln x − ∫(x²/2)(1/x) dx

   = (x²/2) ln x − (1/2)∫x dx

   = (x²/2) ln x − x²/4 + C

   = (x²/4)(2 ln x − 1) + C

7. Problem Solving

   Q7 — Integration by Parts: repeated application

   Find ∫x²e^x dx using integration by parts twice.

   ▶ Show Answer

   First application: Let u = x², dv = e^x dx.   Then du = 2x dx, v = e^x.

   ∫x²e^x dx = x²e^x − ∫2xe^x dx

   Second application for ∫2xe^x dx: Let u = 2x, dv = e^x dx.   Then du = 2 dx, v = e^x.

   ∫2xe^x dx = 2xe^x − 2e^x + C

   Combining: ∫x²e^x dx = x²e^x − (2xe^x − 2e^x) + C

   = e^x(x² − 2x + 2) + C

8. Fluency

   Q8 — Partial Fractions: distinct linear factors

   Find ∫ 5 / [(x + 1)(x − 4)] dx by first expressing as partial fractions.

   ▶ Show Answer

   Write: 5 / [(x+1)(x−4)] = A/(x+1) + B/(x−4)

   Multiply both sides by (x+1)(x−4): 5 = A(x−4) + B(x+1)

   Let x = 4: 5 = 5B ⇒ B = 1

   Let x = −1: 5 = −5A ⇒ A = −1

   ∫ 5/[(x+1)(x−4)] dx = ∫ [−1/(x+1) + 1/(x−4)] dx

   = −ln|x + 1| + ln|x − 4| + C = ln|(x − 4)/(x + 1)| + C

9. Understanding

   Q9 — Partial Fractions: repeated linear factor

   Find ∫ (3x + 1) / [x²(x − 1)] dx.

   ▶ Show Answer

   Write: (3x + 1)/[x²(x − 1)] = A/x + B/x² + C/(x−1)

   Multiply: 3x + 1 = Ax(x−1) + B(x−1) + Cx²

   Let x = 0: 1 = B(0−1) ⇒ B = −1

   Let x = 1: 4 = C(1) ⇒ C = 4

   Expand and compare x² coefficients: 0 = A + C ⇒ A = −4

   Check x coefficient: 3 = −A + B = 4 − 1 = 3 ✓

   ∫ [−4/x − 1/x² + 4/(x−1)] dx

   = −4 ln|x| + 1/x + 4 ln|x−1| + C = 4 ln|(x−1)/x| + 1/x + C

10. Understanding

    Q10 — Partial Fractions: irreducible quadratic factor

    Find ∫ (x + 3) / [(x² + 1)(x − 2)] dx.

    ▶ Show Answer

    Write: (x + 3)/[(x²+1)(x−2)] = (Ax + B)/(x²+1) + C/(x−2)

    Multiply: x + 3 = (Ax + B)(x−2) + C(x²+1)

    Let x = 2: 5 = C(5) ⇒ C = 1

    Expand: x + 3 = Ax² − 2Ax + Bx − 2B + x² + 1

    Compare x²: 0 = A + 1 ⇒ A = −1

    Compare constants: 3 = −2B + 1 ⇒ B = −1

    Compare x: 1 = −2A + B = 2 − 1 = 1 ✓

    ∫ [(−x − 1)/(x²+1) + 1/(x−2)] dx

    = ∫ [−x/(x²+1) − 1/(x²+1) + 1/(x−2)] dx

    = −(1/2)ln(x²+1) − arctan(x) + ln|x−2| + C

11. Fluency

    Q11 — Trig Identities: even power reduction

    Find ∫sin²(x) dx using the identity sin²(x) = (1 − cos 2x)/2.

    ▶ Show Answer

    sin²(x) = (1 − cos 2x)/2

    ∫sin²(x) dx = ∫ (1 − cos 2x)/2 dx = (1/2)∫(1 − cos 2x) dx

    = (1/2)[x − (1/2)sin 2x] + C

    = x/2 − (1/4)sin 2x + C

12. Understanding

    Q12 — Trig Identities: odd power with substitution

    Find ∫cos³(x) dx. Hint: write cos³(x) = cos²(x)cos(x) = (1 − sin²(x))cos(x).

    ▶ Show Answer

    cos³(x) = (1 − sin²(x))cos(x)

    ∫cos³(x) dx = ∫(1 − sin²(x))cos(x) dx

    Let u = sin x.   Then du = cos x dx.

    = ∫(1 − u²) du = u − u³/3 + C

    = sin x − sin³(x)/3 + C

13. Understanding

    Q13 — Trig Identities: product-to-sum formula

    Find ∫sin(3x)cos(x) dx using the product-to-sum identity: sin A cos B = (1/2)[sin(A+B) + sin(A−B)].

    ▶ Show Answer

    Apply identity with A = 3x, B = x:

    sin(3x)cos(x) = (1/2)[sin(4x) + sin(2x)]

    ∫sin(3x)cos(x) dx = (1/2)∫[sin(4x) + sin(2x)] dx

    = (1/2)[−(1/4)cos(4x) − (1/2)cos(2x)] + C

    = −(1/8)cos(4x) − (1/4)cos(2x) + C

14. Problem Solving

    Q14 — Mixed: choose the right technique

    Find ∫ (x² + x + 1) / (x(x + 1)) dx. Determine whether partial fractions or substitution is more efficient.

    ▶ Show Answer

    Partial fractions is the appropriate method (rational function with factorable denominator).

    Write: (x² + x + 1)/[x(x+1)] — degree of numerator equals degree of denominator, so perform polynomial long division first.

    Divide: (x² + x + 1) ÷ (x² + x) = 1 remainder 1.

    So: (x² + x + 1)/[x(x+1)] = 1 + 1/[x(x+1)]

    Now decompose 1/[x(x+1)] = A/x + B/(x+1):

    1 = A(x+1) + Bx.   x = 0: A = 1.   x = −1: B = −1.

    ∫ [1 + 1/x − 1/(x+1)] dx

    = x + ln|x| − ln|x+1| + C = x + ln|x/(x+1)| + C

15. Problem Solving

    Q15 — Mixed: trig integral requiring substitution and identity

    Evaluate ∫₀^π/4 sin(2x) / (1 + cos²(x)) dx. Hint: use sin(2x) = 2 sin x cos x and substitute u = cos x.

    ▶ Show Answer

    Write: sin(2x) = 2 sin x cos x, so the integral becomes:

    ∫₀^π/4 2 sin x cos x / (1 + cos²(x)) dx

    Let u = cos x.   Then du = −sin x dx, so sin x dx = −du.

    Change limits: x = 0 → u = 1;   x = π/4 → u = cos(π/4) = 1/√2.

    ∫₁^1/√2 2u / (1 + u²) × (−du) = ∫_1/√2¹ 2u/(1+u²) du

    Note: d/du[ln(1+u²)] = 2u/(1+u²), so this integral = [ln(1+u²)]_1/√2¹

    = ln(1+1) − ln(1 + 1/2) = ln 2 − ln(3/2)

    = ln(4/3) ≈ 0.288