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Reduction formulaeEdexcel International A Level Further Maths: Revision notes

Section 1

What a reduction formula does

A reduction formula links an integral InI_n to a similar integral with a smaller index, such as In−2I_{n-2}. Applying it repeatedly reduces InI_n to a base case that you can evaluate directly. The two families on the specification are In=∫0π2sin⁡nx dxI_n=\int_0^{\frac{\pi}{2}}\sin^nx\,dx and In=∫sin⁡nxsin⁡x dxI_n=\int\frac{\sin nx}{\sin x}\,dx. In each case you must be able to derive the formula (usually a 'show that' or 'prove' question) and then use it.

Key termsreduction formulabase case
Exam tip

Check which values of nn the formula is valid for. The base case depends on whether nn is odd or even.

Section 2

Deriving nIn=(n−1)In−2nI_n=(n-1)I_{n-2}

Write sin⁡nx=sin⁡n−1xsin⁡x\sin^nx=\sin^{n-1}x\sin x and integrate by parts with u=sin⁡n−1xu=\sin^{n-1}x, dvdx=sin⁡x\frac{dv}{dx}=\sin x: In=[−sin⁡n−1xcos⁡x]0π2+(n−1)∫0π2sin⁡n−2xcos⁡2x dx.I_n=\left[-\sin^{n-1}x\cos x\right]_0^{\frac{\pi}{2}}+(n-1)\int_0^{\frac{\pi}{2}}\sin^{n-2}x\cos^2x\,dx. The boundary term is 00 for n≥2n\ge2 (cosine is 00 at π2\frac{\pi}{2} and sin⁡n−10=0\sin^{n-1}0=0). Replace cos⁡2x\cos^2x by 1−sin⁡2x1-\sin^2x: In=(n−1)(In−2−In) ⇒ nIn=(n−1)In−2.I_n=(n-1)(I_{n-2}-I_n)\ \Rightarrow\ nI_n=(n-1)I_{n-2}.

Key termsintegration by partsboundary term
Common mistake

Forgetting to show the boundary term is zero, or leaving cos⁡2x\cos^2x in the integral instead of turning it into 1−sin⁡2x1-\sin^2x.

Exam tip

Collect the two InI_n terms on the same side: In+(n−1)In=nInI_n+(n-1)I_n=nI_n.

Section 3

Using the formula

Rewrite as In=n−1nIn−2I_n=\frac{n-1}{n}I_{n-2} and work down to the base case. The base cases are I0=∫0π21 dx=π2I_0=\int_0^{\frac{\pi}{2}}1\,dx=\frac{\pi}{2} and I1=∫0π2sin⁡x dx=1I_1=\int_0^{\frac{\pi}{2}}\sin x\,dx=1. Even nn ends at I0I_0: I4=34I2=34×12×π2=3π16I_4=\frac34I_2=\frac34\times\frac12\times\frac{\pi}{2}=\frac{3\pi}{16}. Odd nn ends at I1I_1: I5=45I3=45×23×1=815I_5=\frac45I_3=\frac45\times\frac23\times1=\frac{8}{15}. To use it on other integrals, rewrite them in terms of InI_n. For example ∫0π2sin⁡5xcos⁡2x dx=I5−I7=8105\int_0^{\frac{\pi}{2}}\sin^5x\cos^2x\,dx=I_5-I_7=\frac{8}{105}, and the volume formed by rotating y=sin⁡3xy=\sin^3x about the xx-axis is πI6\pi I_6.

Key termsbase case
Common mistake

Using I1=π2I_1=\frac{\pi}{2} or I0=1I_0=1. They are the other way round.

Exam tip

Write the chain of fractions first, then multiply by the base case at the end.

Section 4

The family ∫sin⁡nxsin⁡x dx\int\frac{\sin nx}{\sin x}\,dx

Let In=∫sin⁡nxsin⁡x dxI_n=\int\frac{\sin nx}{\sin x}\,dx. Use sin⁡(n+2)x−sin⁡nx=2cos⁡(n+1)xsin⁡x\sin(n+2)x-\sin nx=2\cos(n+1)x\sin x, found by expanding sin⁡((n+1)x±x)\sin\big((n+1)x\pm x\big) with the compound-angle formulae. Dividing by sin⁡x\sin x and integrating gives In+2=In+2sin⁡(n+1)xn+1.I_{n+2}=I_n+\frac{2\sin(n+1)x}{n+1}. Base cases: I1=∫1 dx=xI_1=\int1\,dx=x and I2=∫2cos⁡x dx=2sin⁡xI_2=\int2\cos x\,dx=2\sin x. So I3=x+sin⁡2xI_3=x+\sin2x and I5=x+sin⁡2x+12sin⁡4xI_5=x+\sin2x+\frac12\sin4x. Here the index steps up by 22 and the result contains trigonometric terms, not just numbers.

Key termscompound-angle formula
Exam tip

The sin⁡(n+1)xcos⁡x\sin(n+1)x\cos x terms cancel when you subtract the two expansions; the cos⁡(n+1)xsin⁡x\cos(n+1)x\sin x terms double.

Common mistake

Integrating cos⁡(n+1)x\cos(n+1)x without dividing by n+1n+1.

Section 5

Exam technique

  • 'Show that' or 'prove': every step must appear, including the boundary term.
  • Quote the relation you are using with its value of nn each time (for example 5I5=4I35I_5=4I_3).
  • Keep exact values (π\pi and fractions); do not use decimals unless asked.
  • Link the integral in the question to InI_n before applying the formula, using cos⁡2x=1−sin⁡2x\cos^2x=1-\sin^2x where needed.
Exam tip

Sense check: 0≤sin⁡nx≤10\le\sin^nx\le1 on [0,π2]\left[0,\frac{\pi}{2}\right], so each InI_n must be less than π2\frac{\pi}{2}.

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Exam questions on Reduction formulae

  1. Let In=∫0π2sin⁡nx dxI_n=\int_0^{\frac{\pi}{2}}\sin^n x\,dx for integers n≥0n\ge0. It is given that nIn=(n−1)In−2nI_n=(n-1)I_{n-2} for n≥2n\ge2.
    Hence find the exact value of I4I_4.2 marks
  2. The region RR lies between the curve y=sin⁡3xy=\sin^3x, the xx-axis and the line x=π2x=\frac{\pi}{2}, for 0≤x≤π20\le x\le\frac{\pi}{2}. Let In=∫0π2sin⁡nx dxI_n=\int_0^{\frac{\pi}{2}}\sin^n x\,dx, where nIn=(n−1)In−2nI_n=(n-1)I_{n-2} for n≥2n\ge2.
    Find the exact area of RR.2 marks
  3. For integers n≥1n\ge1, let In=∫sin⁡nxsin⁡x dxI_n=\int\frac{\sin nx}{\sin x}\,dx, where sin⁡x≠0\sin x\ne0. Constants of integration may be ignored.
    Show that sin⁡(n+2)x−sin⁡nx=2cos⁡(n+1)xsin⁡x\sin(n+2)x-\sin nx=2\cos(n+1)x\sin x.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).