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 a whole family falls out 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 that can be integrated directly.
- Recognise when a formula drops by one step or by two.
WHAT YOU PROBABLY THINK
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 gives a reduction formula. One integration by parts serves every n at once, so the opening claim overcounts by a factor of 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, odd powers lose it, because the two chains start from different base cases.
Deriving one from scratch
Exam questions usually ask for the derivation, so the by-parts step must be visible. Choose the split carefully: the factor that will be differentiated should get simpler, and the factor integrated 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, rather than by two, because only one factor of x is lost per integration.
YOUR TURN
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: xn is tiny across most of [0, 1] for large n, so the integral must fall, and the alternating-looking algebra still produces positive numbers.
THE EXAM BIT
- State I sub n as an integral before deriving anything; the notation is half the method mark.
- Show the by-parts line in full, including the evaluated bracket, before rearranging.
- 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
3
I
₃
=
2
I
₁
=
2
,
s
o
I
₃
=
2
/
3
.
Integrate by parts once, rearrange, and the result is a recurrence linking I sub n to a smaller case.
Chase the recurrence down to a base case you can integrate directly; even and odd n often behave differently.
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.
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 that can be integrated directly.
- Recognise when a formula drops by one step or by two.
Open the full revision checklist to see every objective in the course in one place.
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