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Integration by substitution

A substitution changes an awkward integral in x into a simpler one in u. In harder questions the substitution is given.

The method

1. Choose u (or use the one given), for example u = x - 1.

2. Find du/dx and rearrange to replace dx: dx = du/(du/dx).

3. Write everything in terms of u, including any leftover x.

4. Integrate with respect to u.

5. Substitute back to x (indefinite) or use u-limits (definite).

Changing limits

For a definite integral, change the x-limits into u-limits using the substitution. Then you never need to return to x. With u = 2x - 1, x = 1 and x = 5 become u = 1 and u = 9.

Trig substitutions

Some integrals need a substitution such as x = sin θ. Then dx = cos θ dθ, and sqrt(1 - x2) = cos θ. Expect these to be given in the question.

Worked example

Use u = x - 1 to find ∫1^{2} x(x - 1)4 dx.

  1. du = dx; x = u + 1; limits u = 0 to 1
  2. ∫0^{1} (u + 1)u4 du = ∫0^{1} (u5 + u4) du
  3. = 1/6 + 1/5

Answer: 11/30

Key idea: Replace dx using du/dx, write everything in u, change the limits, integrate.

Check yourself

  1. Using u = x2 + 3, what does ∫ x sqrt(x2 + 3) dx become?
    • ∫ sqrt(u) du
    • 2∫ sqrt(u) du
    • (1/2)∫ sqrt(u) du
    • ∫ x sqrt(u) du
    Show answer

    (1/2)∫ sqrt(u) du. du/dx = 2x, so x dx = (1/2)du.

  2. Use u = x + 1 to evaluate ∫0^{3} x sqrt(x + 1) dx. Give your answer as a fraction in its simplest form.
    Show answer

    116/15. x = u - 1, dx = du, limits 1 to 4. ∫1^{4} (u3/2 - u1/2) du = (2/5)u5/2 - (2/3)u3/2. From 1 to 4: (64/5 - 16/3) - (2/5 - 2/3) = 116/15.

  3. Using u = 2x - 1, what are the new limits for ∫1^{5} ... dx?
    • u = 1 and u = 5
    • u = 0 and u = 4
    • u = 1 and u = 9
    • u = 3 and u = 11
    Show answer

    u = 1 and u = 9. x = 1 gives u = 1; x = 5 gives u = 9.

  4. Use u = cos x to evaluate ∫0^{π/2} sin x cos2 x dx. Give your answer as a fraction in its simplest form.
    Show answer

    1/3. du = -sin x dx; limits u = 1 to 0. -∫1^{0} u2 du = ∫0^{1} u2 du = 1/3.

  5. Use u = sqrt(x) to evaluate ∫0^{4} 1/(1 + sqrt(x)) dx, to 3 decimal places.
    Show answer

    1.803. x = u2, dx = 2u du, limits 0 to 2. ∫0^{2} 2u/(1 + u) du = ∫0^{2} (2 - 2/(1 + u)) du = 4 - 2ln 3 = 1.803.

  6. Using u = x2 + 1, what should dx be replaced with?
    • du
    • du/(2x)
    • 2x du
    • du/x2
    Show answer

    du/(2x). du/dx = 2x, so dx = du/(2x).

  7. Use x = sin θ to evaluate ∫0^{1} sqrt(1 - x2) dx, to 3 decimal places.
    Show answer

    0.785. dx = cos θ dθ, limits 0 to π/2: ∫0^{π/2} cos2 θ dθ = θ/2 + (1/4)sin 2θ, evaluated from 0 to π/2, = π/4 = 0.785.

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Last updated 7 November 2026. Spotted a mistake? Tell us.