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Reduction formulaeEdexcel A-Level Further Maths: Revision notes

Section 1

Why reduction formulae?

Some integrals depend on an integer nn, such as ∫xnex dx\int x^ne^x\,dx or ∫sin⁡nx dx\int\sin^nx\,dx. A reduction formula links the integral InI_n to an integral with a smaller index, such as In−1I_{n-1} or In−2I_{n-2}. Applying it repeatedly reduces the problem to a base case (I0I_0 or I1I_1) that is easy to integrate directly. The two steps are always: derive the formula (usually by parts), then use it to evaluate a particular InI_n by working down to the base case.

Key termsreduction formulabase case
Exam tip

Evaluate the base case first and check it by integrating directly, for example I0=∫0π21 dx=π2I_0=\int_0^{\frac{\pi}{2}}1\,dx=\frac{\pi}{2}.

Section 2

Deriving nIn=(n−1)In−2nI_n=(n-1)I_{n-2} for sin⁡nx\sin^nx

Let In=∫0π2sin⁡nx dxI_n=\int_0^{\frac{\pi}{2}}\sin^nx\,dx. 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, v′=sin⁡xv'=\sin x: In=[−sin⁡n−1xcos⁡x]0π2+(n−1)∫0π2sin⁡n−2xcos⁡2x dx.I_n=[-\sin^{n-1}x\cos x]_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\geq2. Replace cos⁡2x=1−sin⁡2x\cos^2x=1-\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}. With I0=π2I_0=\frac{\pi}{2} and I1=1I_1=1: I3=23I_3=\frac23, I4=34⋅12⋅π2=3π16I_4=\frac34\cdot\frac12\cdot\frac{\pi}{2}=\frac{3\pi}{16}, I5=815I_5=\frac{8}{15}.

Key termsintegration by partsboundary term
Common mistake

Forgetting that ddxsin⁡n−1x=(n−1)sin⁡n−2xcos⁡x\frac{d}{dx}\sin^{n-1}x=(n-1)\sin^{n-2}x\cos x: the factor (n−1)(n-1) is easily lost.

Exam tip

Odd nn ends at I1=1I_1=1; even nn ends at I0=π2I_0=\frac{\pi}{2}.

Section 3

Reduction by parts: xnexx^ne^x and xnsin⁡xx^n\sin x

Where the power of xx falls by one on differentiating, integrate by parts with u=xnu=x^n. For Jn=∫01xnex dxJ_n=\int_0^1x^ne^x\,dx: Jn=[xnex]01−nJn−1=e−nJn−1J_n=[x^ne^x]_0^1-nJ_{n-1}=e-nJ_{n-1}. Since J0=e−1J_0=e-1: J1=1J_1=1, J2=e−2J_2=e-2, J3=6−2eJ_3=6-2e. For Kn=∫0π2xnsin⁡x dxK_n=\int_0^{\frac{\pi}{2}}x^n\sin x\,dx the formula needs two applications of parts (sine to cosine and back to sine): Kn=n(π2)n−1−n(n−1)Kn−2K_n=n\left(\frac{\pi}{2}\right)^{n-1}-n(n-1)K_{n-2} for n≥2n\geq2. With K0=1K_0=1: K2=π−2K_2=\pi-2.

Key termstwo-step reduction
Common mistake

Reusing a boundary term from another integral: it changes if the limits change, so recompute it each time.

Section 4

Trigonometric reductions using identities

Not every reduction uses parts. For Tn=∫0π4tan⁡nx dxT_n=\int_0^{\frac{\pi}{4}}\tan^nx\,dx use tan⁡2x+1=sec⁡2x\tan^2x+1=\sec^2x: Tn+Tn−2=∫0π4tan⁡n−2xsec⁡2x dx=[tan⁡n−1xn−1]0π4=1n−1.T_n+T_{n-2}=\int_0^{\frac{\pi}{4}}\tan^{n-2}x\sec^2x\,dx=\left[\frac{\tan^{n-1}x}{n-1}\right]_0^{\frac{\pi}{4}}=\frac{1}{n-1}. So Tn=1n−1−Tn−2T_n=\frac{1}{n-1}-T_{n-2}, with T0=π4T_0=\frac{\pi}{4}. For example T2=1−π4T_2=1-\frac{\pi}{4} and T4=π4−23T_4=\frac{\pi}{4}-\frac23.

Key termstan-sec identity
Exam tip

If a power of tan⁡x\tan x multiplies sec⁡2x\sec^2x, substitute u=tan⁡xu=\tan x.

Section 5

The formula In+2=In+2sin⁡((n+1)x)n+1I_{n+2}=I_n+\frac{2\sin((n+1)x)}{n+1}

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 (from the factor formulae, or by expanding sin⁡((n+1)x±x)\sin((n+1)x\pm x)): In+2−In=∫sin⁡(n+2)x−sin⁡nxsin⁡x dx=∫2cos⁡(n+1)x dx=2sin⁡(n+1)xn+1+c.I_{n+2}-I_n=\int\frac{\sin(n+2)x-\sin nx}{\sin x}\,dx=\int2\cos(n+1)x\,dx=\frac{2\sin(n+1)x}{n+1}+c. So In+2=In+2sin⁡((n+1)x)n+1I_{n+2}=I_n+\frac{2\sin((n+1)x)}{n+1}, as an indefinite integral (the constant is absorbed by the chosen limits). Starting from I1=∫1 dx=xI_1=\int1\,dx=x gives I3=x+sin⁡2xI_3=x+\sin2x.

Key termsfactor formula
Common mistake

Dividing by nn instead of n+1n+1 when integrating cos⁡(n+1)x\cos(n+1)x.

Section 6

Using a reduction formula in an exam

  1. State the base case(s) and evaluate them directly.
  2. Use the formula with consecutive values of nn, keeping answers exact (leave π\pi and ee in the answer).
  3. If asked to show a formula, write each integration by parts in full, including the boundary term, and state why it vanishes.
  4. Check a value numerically if you can: ∫01x3ex dx=6−2e≈0.563\int_0^1x^3e^x\,dx=6-2e\approx0.563 is positive and smaller than ∫01ex dx=e−1≈1.72\int_0^1e^x\,dx=e-1\approx1.72.
Exam tip

Write the formula with the actual value of nn substituted before using it: 4I4=3I24I_4=3I_2, not a rearranged form you then misuse.

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Carry on to the next subtopic.

Exam questions on Reduction formulae

  1. For integers n≥0n\geq0, let In=∫0π2sin⁡nx dxI_n=\int_0^{\frac{\pi}{2}}\sin^n x\,dx. It is known that nIn=(n−1)In−2nI_n=(n-1)I_{n-2} for n≥2n\geq2.
    Find the exact value of I5I_5.2 marks
  2. For integers n≥0n\geq0, let Jn=∫01xnex dxJ_n=\int_0^1 x^n e^x\,dx.
    Use the result of part (a) to find the exact value of J3J_3.2 marks
  3. For integers n≥0n\geq0, let Tn=∫0π4tan⁡nx dxT_n=\int_0^{\frac{\pi}{4}}\tan^n x\,dx.
    Show that Tn+Tn−2=1n−1T_n+T_{n-2}=\frac{1}{n-1} for n≥2n\geq2.3 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).