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Reduction formulaeEdexcel A-Level Further Maths: Flashcards

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Question

State the reduction formula for $I_n=\int_0^{\frac{\pi}{2}}\sin^nx\,dx$.

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State the reduction formula for In=∫0π2sin⁡nx dxI_n=\int_0^{\frac{\pi}{2}}\sin^nx\,dx.
nIn=(n−1)In−2nI_n=(n-1)I_{n-2} for n≥2n\geq2.
Values of I0I_0 and I1I_1 for ∫0π2sin⁡nx dx\int_0^{\frac{\pi}{2}}\sin^nx\,dx?
I0=π2I_0=\frac{\pi}{2} and I1=1I_1=1.
Which method derives nIn=(n−1)In−2nI_n=(n-1)I_{n-2}?
Integration by parts with u=sin⁡n−1xu=\sin^{n-1}x, v′=sin⁡xv'=\sin x, then cos⁡2x=1−sin⁡2x\cos^2x=1-\sin^2x.
Find ∫0π2sin⁡3x dx\int_0^{\frac{\pi}{2}}\sin^3x\,dx.
23\frac23 (from 3I3=2I13I_3=2I_1).
Find ∫0π2sin⁡4x dx\int_0^{\frac{\pi}{2}}\sin^4x\,dx.
3π16\frac{3\pi}{16}.
Reduction formula for Jn=∫01xnex dxJ_n=\int_0^1x^ne^x\,dx?
Jn=e−nJn−1J_n=e-nJ_{n-1}.
What is J0J_0 and hence J1J_1?
J0=e−1J_0=e-1, J1=1J_1=1.
Why does ∫0π2xnsin⁡x dx\int_0^{\frac{\pi}{2}}x^n\sin x\,dx need integration by parts twice?
One application turns sin⁡x\sin x into cos⁡x\cos x; a second turns it back to sin⁡x\sin x with xn−2x^{n-2}.
Result for Kn=∫0π2xnsin⁡x dxK_n=\int_0^{\frac{\pi}{2}}x^n\sin x\,dx?
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}, n≥2n\geq2.
Relation between TnT_n and Tn−2T_{n-2} for Tn=∫0π4tan⁡nx dxT_n=\int_0^{\frac{\pi}{4}}\tan^nx\,dx?
Tn+Tn−2=1n−1T_n+T_{n-2}=\frac{1}{n-1}.
Identity behind the reduction for ∫sin⁡nxsin⁡x dx\int\frac{\sin nx}{\sin x}\,dx?
sin⁡(n+2)x−sin⁡nx=2cos⁡(n+1)xsin⁡x\sin(n+2)x-\sin nx=2\cos(n+1)x\sin x.
Reduction formula for In=∫sin⁡nxsin⁡x dxI_n=\int\frac{\sin nx}{\sin x}\,dx?
In+2=In+2sin⁡((n+1)x)n+1I_{n+2}=I_n+\frac{2\sin((n+1)x)}{n+1} (plus a constant).

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).