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Substitutions and quadratic surds

Edexcel IAL Further Mathematics · WFM03 Further Pure Mathematics 3

Integrals containing sqrt(a2 - x2), sqrt(x2 + a2) or sqrt(x2 - a2) are tamed by a trigonometric or hyperbolic substitution that turns the surd into a single term. Quadratic surds like sqrt(x2 + 6x + 13) need completing the square first.

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Which substitution?

The specification says that in more complicated cases the substitution will be given, so for simple ones you may need to choose it yourself.

Carrying it through

1. Write dx in terms of du.

2. Change the limits (or change back to x at the end).

3. Simplify; you often get cos2 u or cosh2 u, so use a double angle identity.

4. For the answer, sinh 2u = 2 sinh u cosh u lets you go back to x: sinh u = x/a and cosh u = sqrt(1 + x2/a2).

Quadratic surds

Complete the square: x2 + 6x + 13 = (x + 3)2 + 4. Then substitute x + 3 = 2 sinh u, or use the standard result directly: ∫ 1/sqrt((x + 3)2 + 4) dx = arsinh((x + 3)/2) + c.

If x2 has a coefficient: 4x2 - 4x + 5 = (2x - 1)2 + 4.

Worked example

Evaluate ∫ from 0 to 2 of sqrt(x2 + 4) dx using x = 2 sinh u.

  1. dx = 2 cosh u du and sqrt(4 sinh2 u + 4) = 2 cosh u, so the integrand is 4cosh2 u = 2(1 + cosh 2u)
  2. Limits: x = 0 gives u = 0; x = 2 gives sinh u = 1, u = ln(1 + sqrt(2))
  3. [2u + sinh 2u] = [2u + 2 sinh u cosh u], and at the top sinh u = 1, cosh u = sqrt(2)

Answer: 2 ln(1 + sqrt(2)) + 2sqrt(2), about 4.591

Key idea: sqrt(a2 - x2): x = a sin u. sqrt(x2 + a2): x = a sinh u. sqrt(x2 - a2): x = a cosh u. Change dx and the limits; complete the square for quadratic surds.

Check yourself

  1. Use x = sin u to evaluate ∫ from 0 to 1 of sqrt(1 - x2) dx, to 3 decimal places.
    Show answer

    0.785. ∫ from 0 to π/2 of cos2 u du = π/4 = 0.785 (a quarter of the unit circle).

  2. Use x = 3 sinh u to evaluate ∫ from 0 to 3 of sqrt(9 + x2) dx, to 3 decimal places.
    Show answer

    10.330. The integrand becomes 9cosh2 u, giving (9/2)[u + sinh u cosh u] from 0 to arsinh 1 = (9/2)(ln(1 + sqrt(2)) + sqrt(2)) = 10.330.

  3. Under x = 2 sinh u, what does sqrt(x2 + 4) become?
    • 2 sinh u
    • 4 cosh u
    • 2 cosh u
    • cosh u + 2
    Show answer

    2 cosh u. sqrt(4 sinh2 u + 4) = sqrt(4 cosh2 u) = 2 cosh u.

  4. How should you start ∫ 1/sqrt(x2 + 6x + 13) dx?
    • Write x2 + 6x + 13 as (x + 3)2 - 4
    • Write x2 + 6x + 13 as (x + 3)2 + 4
    • Factorise x2 + 6x + 13
    • Substitute u = x2 + 6x + 13
    Show answer

    Write x2 + 6x + 13 as (x + 3)2 + 4. (x + 3)2 = x2 + 6x + 9, and 13 - 9 = 4. The integral is then arsinh((x + 3)/2) + c.

  5. Evaluate ∫ from -3 to 1 of 1/sqrt(x2 + 6x + 13) dx, to 3 decimal places.
    Show answer

    1.444. [arsinh((x + 3)/2)] from -3 to 1 = arsinh 2 - arsinh 0 = ln(2 + sqrt(5)) = 1.444.

  6. Use x = 3 sin u to evaluate ∫ from 0 to 3/2 of 1/(9 - x2)3/2 dx, to 4 decimal places.
    Show answer

    0.0642. The integrand becomes 3 cos u/(27 cos3 u) = (1/9) sec2 u, giving (1/9)[tan u] from 0 to π/6 = 1/(9sqrt(3)) = 0.0642.

  7. Evaluate ∫ from 1/2 to 3/2 of 1/sqrt(4x2 - 4x + 5) dx, to 3 decimal places.
    Show answer

    0.441. 4x2 - 4x + 5 = (2x - 1)2 + 4. With u = 2x - 1 the integral is (1/2)[arsinh(u/2)] from 0 to 2 = (1/2) arsinh 1 = 0.441.

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Last updated 8 November 2026. Spotted a mistake? Tell us.