Practice Maths

← Integration Techniques › Integration by Substitution › Solutions

Integration by Substitution — Full Worked Solutions

  1. Find ∫2x(x² + 3)&sup4; dx using u = x² + 3. Fluency

    Let u = x² + 3.   Differentiate: du/dx = 2x, so du = 2x dx.

    Substitute: ∫2x(x² + 3)&sup4; dx = ∫u&sup4; du

    = u&sup5;/5 + C

    Back-substitute: = (x² + 3)&sup5; / 5 + C

    Check by differentiating: d/dx[(x²+3)&sup5;/5] = (1/5) × 5(x²+3)&sup4; × 2x = 2x(x²+3)&sup4; ✓

  2. Find ∫6x²(x³ + 1)² dx. Fluency

    Let u = x³ + 1.   Differentiate: du/dx = 3x², so du = 3x² dx, which means 6x² dx = 2 du.

    Substitute: ∫6x²(x³ + 1)² dx = 2∫u² du

    = 2 × u³/3 + C = 2u³/3 + C

    Back-substitute: = 2(x³ + 1)³ / 3 + C

  3. Find ∫cos(2x) dx. Fluency

    Let u = 2x.   Differentiate: du/dx = 2, so du = 2 dx, meaning dx = du/2.

    Substitute: ∫cos(2x) dx = ∫cos(u) × du/2 = (1/2)∫cos(u) du

    = (1/2)sin(u) + C

    Back-substitute: = (1/2)sin(2x) + C

  4. Find ∫(3x + 1)&sup5; dx. Fluency

    Let u = 3x + 1.   Differentiate: du/dx = 3, so du = 3 dx, meaning dx = du/3.

    Substitute: ∫(3x + 1)&sup5; dx = ∫u&sup5; × du/3 = (1/3)∫u&sup5; du

    = (1/3) × u&sup6;/6 + C = u&sup6;/18 + C

    Back-substitute: = (3x + 1)&sup6; / 18 + C

  5. Evaluate ∫01 2xex² dx. Understanding

    Let u = x².   Differentiate: du/dx = 2x, so du = 2x dx.

    Change limits: x = 0 → u = 0² = 0;   x = 1 → u = 1² = 1.

    Substitute: ∫01 2xex² dx = ∫01 eu du

    = [eu]01 = e1 − e0 = e − 1

    ≈ 1.718

  6. Find ∫ x / √(x² + 4) dx. Understanding

    Let u = x² + 4.   Differentiate: du/dx = 2x, so du = 2x dx, meaning x dx = du/2.

    Substitute: ∫ x/√(x²+4) dx = ∫ (1/√u) × du/2 = (1/2)∫u−1/2 du

    = (1/2) × 2u1/2 + C = u1/2 + C

    Back-substitute: = √(x² + 4) + C

    Check: d/dx[√(x²+4)] = (1/2)(x²+4)−1/2 × 2x = x/√(x²+4) ✓

  7. Find ∫sin²(x)cos(x) dx using u = sin x. Understanding

    Let u = sin x.   Differentiate: du/dx = cos x, so du = cos x dx.

    Substitute: ∫sin²(x)cos(x) dx = ∫u² du

    = u³/3 + C

    Back-substitute: = sin³(x) / 3 + C

  8. Find ∫(ln x) / x dx. Understanding

    Let u = ln x.   Differentiate: du/dx = 1/x, so du = dx/x.

    Note: (ln x)/x dx = ln x × (dx/x) = u du.

    Substitute: ∫(ln x)/x dx = ∫u du

    = u²/2 + C

    Back-substitute: = (ln x)² / 2 + C

  9. Evaluate ∫0π/4 tan x dx. Problem Solving

    Write tan x = sin x / cos x.

    Let u = cos x.   Differentiate: du/dx = −sin x, so du = −sin x dx, meaning sin x dx = −du.

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

    Substitute: ∫0π/4 sin x/cos x dx = ∫11/√2 (1/u)(−du)

    = −[ln|u|]11/√2 = −(ln(1/√2) − ln 1) = −ln(1/√2)

    = ln(√2) = (1/2)ln 2

    = (1/2)ln 2 ≈ 0.347

  10. A curve has dy/dx = 3x²ex³. Find y given y = 2 when x = 0. Problem Solving

    Integrate both sides: y = ∫3x²ex³ dx.

    Let u = x³.   Differentiate: du/dx = 3x², so du = 3x² dx.

    Substitute: y = ∫eu du = eu + C = ex³ + C

    Apply initial condition: y = 2 when x = 0.

    2 = e0³ + C = e0 + C = 1 + C

    C = 1

    y = ex³ + 1