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For all $n \in \mathbb{N} : n≥2$, I might add. $$I_n:=\int_{0}^{\pi}(\sin x)^ndx$$ $$I_n=\frac{n-1}{n}I_{n-2}$$

I've tried to rewrite $\int(\sin x)^ndx$ to the form $\int(\sin x)(\sin x)^{n-1}dx$ but then I don't get much further. Any advice on how to move forward would be appreciated.

S.C.B.
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Void
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    See http://math.stackexchange.com/questions/236543/proving-a-reduction-formula-for-the-antiderivative-of-cosnx OR http://www.vias.org/calculus/07_trigonometric_functions_05_03.html – lab bhattacharjee Jun 17 '16 at 09:37
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    This has been asked before in the last days... make a search. – mickep Jun 17 '16 at 09:44
  • See http://math.stackexchange.com/questions/1825535/how-do-i-show-the-relationship-between-i-n-int-0-pisinxndx-and-i-n – cgiovanardi Jun 17 '16 at 13:56

2 Answers2

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The initial step is correct; assuming $n>1$, \begin{align} I_n &=\int_0^\pi (\sin x)^{n-1}\sin x\,dx \\ &=\Bigl[-(\sin x)^{n-1}\cos x\Bigr]_0^\pi +(n-1)\int_0^\pi (\sin x)^{n-2}\cos^2x\,dx\\ &=(n-1)\int_0^\pi (\sin x)^{n-2}(1-\sin^2x)\,dx\\ &=(n-1)\int_0^\pi (\sin x)^{n-2}\,dx- (n-1)\int_0^\pi (\sin x)^{n}\,dx\\[6px] &=(n-1)I_{n-2}-(n-1)I_n \end{align}

egreg
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  • Thanks a lot! How do I get the last term to be $I_n=\frac{n-1}{n}I_{n-2}$ exactly? – Void Jun 17 '16 at 10:50
  • @Void Just expand $I_n=(n-1)I_{n-2}-nI_n+I_n$, cancel the terms and switch side to $nI_n$. – egreg Jun 17 '16 at 10:51
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$I_n= \int ^{\pi}_0 \sin^{n-1}x.\sin{x}$ $dx=\sin^{n-1}x.(-\cos{x})-\int ^{\pi}_0 (n-1)\sin^{n-2}x.\cos{x}(-\cos{x})$ $dx$

$=\sin^{n-1}x.(-\cos{x})+\int ^{\pi}_0 (n-1)\sin^{n-2}x.\cos^2{x}$ $dx$

$=\sin^{n-1}x.(-\cos{x})+\int ^{\pi}_0 (n-1)\sin^{n-2}x.(1-\sin^2{x})$ $dx$

$=\sin^{n-1}x.(-\cos{x})|_0^{\pi}+(n-1)\int ^{\pi}_0 (\sin^{n-2}x-\sin^n{x})$ $dx$

$2I_n=0+(n-1)\int ^{\pi}_0 (\sin^{n-2}x)$ $dx=(n-1)I_{n-2}$

$\therefore I_n=\frac{(n-1)}{2}I_{n-2}$

Tosh
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