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Using reduction formulae

Edexcel IAL Further Mathematics · WFM03 Further Pure Mathematics 3

Some reduction formulae need integration by parts twice, or a hyperbolic identity. Exam questions usually ask you to prove the formula, then use it to evaluate a particular integral exactly.

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Parts twice

I_n = ∫ from 0 to π/2 of xⁿ cos x dx.

First: u = xⁿ, v = sin x gives I_n = (π/2)ⁿ - n ∫ x^n-1 sin x dx.

Second: u = x^n-1, v = -cos x; the boundary term is zero, leaving ∫ x^n-1 sin x dx = (n - 1) I_(n-2).

So I_n = (π/2)ⁿ - n(n - 1) I_(n-2). Two rounds of parts bring cos back.

Logs and hyperbolic powers

I_n = ∫ from 1 to e of (ln x)ⁿ dx: parts with u = (ln x)ⁿ, v = x gives I_n = e - n I_(n-1).
I_n = ∫ coshⁿ x dx: parts with u = cosh^n-1 x, v = sinh x and sinh² x = cosh² x - 1 give n I_n = sinh x cosh^n-1 x + (n - 1) I_(n-2).

Note the + sign, where the trigonometric version has -.

Exam technique

Show every step of the proof: the 'show that' mark scheme needs the boundary terms evaluated.
Write down the starting values I_0 and I_1 clearly.
Work down in order and keep exact values (π, e, ln 2) until the end.

Worked example

I_n = ∫ from 0 to π/2 of xⁿ cos x dx. Use I_n = (π/2)ⁿ - n(n - 1) I_(n-2) to find I_2 exactly.

I_0 = ∫ from 0 to π/2 of cos x dx = 1
I_2 = (π/2)² - 2(1) I_0

Answer: π²/4 - 2, about 0.467

Key idea: If one round of parts does not reduce the integral to I_(n-1) or I_(n-2), do a second round. Start values first, exact values throughout.

Check yourself

Using I_n = (π/2)ⁿ - n(n - 1) I_(n-2) and I_2 = π²/4 - 2, find I_4 = ∫ from 0 to π/2 of x⁴ cos x dx, to 3 decimal places.
Show answer
0.479. I_4 = (π/2)⁴ - 12 I_2 = π⁴/16 - 3π² + 24 = 0.479.
Find I_1 = ∫ from 0 to π/2 of x cos x dx, to 3 decimal places.
Show answer
0.571. [x sin x + cos x] from 0 to π/2 = π/2 - 1 = 0.571.
Why does the reduction formula for ∫ xⁿ cos x dx need integration by parts twice?
- The first round increases the power of x
- The boundary terms do not cancel
- cos x cannot be integrated
- After one round the integral contains sin x, so a second round is needed to get back to cos x
Show answer
After one round the integral contains sin x, so a second round is needed to get back to cos x. One round gives ∫ x^n-1 sin x dx, which is not of the form I_m. A second round gives ∫ x^n-2 cos x dx = I_(n-2).
I_n = ∫ from 0 to π/4 of tanⁿ x dx satisfies I_n + I_(n-2) = 1/(n - 1), and I_2 = 1 - π/4. Find I_4, to 3 decimal places.
Show answer
0.119. I_4 = 1/3 - I_2 = π/4 - 2/3 = 0.119.
I_n = ∫ from 1 to e of (ln x)ⁿ dx satisfies I_n = e - n I_(n-1), with I_0 = e - 1. Find I_3, to 3 decimal places.
Show answer
0.563. I_1 = 1, I_2 = e - 2, I_3 = e - 3(e - 2) = 6 - 2e = 0.563.
For I_n = ∫ from 1 to e of (ln x)ⁿ dx, which parts choice gives the reduction formula?
- u = (ln x)ⁿ, dv/dx = 1
- u = 1, dv/dx = (ln x)ⁿ
- u = ln x, dv/dx = (ln x)^n-1
- u = x, dv/dx = (ln x)ⁿ / x
Show answer
u = (ln x)ⁿ, dv/dx = 1. du/dx = n(ln x)^n-1/x, and multiplying by v = x cancels the 1/x, leaving n(ln x)^n-1.
I_n = ∫ from 0 to ln 2 of coshⁿ x dx satisfies n I_n = sinh(ln 2) cosh^n-1(ln 2) + (n - 1) I_(n-2), with I_0 = ln 2. Find I_2, to 3 decimal places.
Show answer
0.815. sinh(ln 2) = 3/4 and cosh(ln 2) = 5/4. 2 I_2 = (3/4)(5/4) + ln 2, so I_2 = 15/32 + (1/2) ln 2 = 0.815.

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Sources

Pearson Edexcel International Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification (issue 3), Unit FP3

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