Maths › Further Pure 2 › Reduction formulae
Reduction formulae
Integrate by parts once and the answer contains a smaller version of the same integral. Turn that into a recurrence and the whole family follows from one base case.
Builds on Integration by substitution and by parts and Summing series.
IN THIS TOPIC
- Derive a reduction formula by integrating by parts and rearranging.
- Apply the recurrence down to a base case you can integrate directly.
- Recognise when a formula drops by one step and when it drops by two.
COMMON MISCONCEPTION
An integral of sin to the tenth power would take ten separate integrations by parts.
Parts, once, then recur
Write In for the integral with parameter n, split the integrand so that one part differentiates towards something simpler, and integrate by parts. What comes back contains In-1 or In-2, and rearranging turns it into a reduction formula. That single integration by parts serves every n at once; sin to the tenth power needs one of them, not ten.
WORKED EXAMPLE
Down the ladder to a base case
Given nIn = (n − 1)In-2 for the sine integrals from 0 to π/2, evaluate I₄ and I₅.
Base cases: I₀ = π/2, since the integrand is 1, and I₁ = 1.
I₂ = (1/2)I₀ = π/4, then I₄ = (3/4)I₂ = 3π/16.
I₃ = (2/3)I₁ = 2/3, then I₅ = (4/5)I₃ = 8/15.
Even powers keep the π and odd powers lose it, because the two chains start from different base cases.
Deriving one from scratch
Exam questions nearly always ask for the derivation, so the by-parts step has to be visible on the page. Choose the split carefully. The factor you differentiate should get simpler, and the factor you integrate should not get worse.
WORKED EXAMPLE
A polynomial against an exponential
For In = ∫xnex dx from 0 to 1, derive a reduction formula and find I₃.
By parts with u = xn and dv = exdx: In = [xnex] − n∫xn-1ex dx, so In = e − nIn-1.
I₀ = e − 1. Then I₁ = e − (e − 1) = 1, I₂ = e − 2 and I₃ = e − 3(e − 2) = 6 − 2e ≈ 0.5634.
Each step drops n by one here, not by two, because only one factor of x is lost per integration.
GUIDED PRACTICE
Reading the formula backwards
Given In = e − nIn-1 with I₀ = e − 1, find I₄ exactly, and comment on the size of the answer.
Show the working
I₃ = 6 − 2e from the previous example.
I₄ = e − 4(6 − 2e) = 9e − 24 ≈ 0.4645.
The values shrink slowly towards zero, because xn is tiny across most of [0, 1] once n is large. The alternating-looking algebra still produces positive numbers, which is a useful check on the arithmetic.
ASSESSMENT FOCUS
- State I sub n as an integral before deriving anything; the notation carries half the method mark.
- Show the by-parts line in full, evaluated bracket included, before you rearrange.
- Identify the base case explicitly, and check whether the recurrence steps down by one or by two.
- Keep answers exact. π and e belong in the final line, not their decimal values.
CHECK YOURSELF
Given nIn = (n − 1)In-2 with I₁ = 1, find I₃.
Show a hint
Put n = 3 into the formula.
Show the answer
3I₃ = 2I₁ = 2, so I₃ = 2/3.
Integrate by parts once, rearrange into a recurrence, then walk it down to a base case you can integrate directly.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the reduction formulae questions page.
CHECK YOUR PROGRESS
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- Derive a reduction formula by integrating by parts and rearranging.
- Apply the recurrence down to a base case you can integrate directly.
- Recognise when a formula drops by one step and when it drops by two.
Open the full revision checklist to see every objective in the course in one place.