If $∫_0^{π/2} \cos ^nxdx=I_n$, prove that $nI_n=\left(n-1\right)I_{n-2}, \left(n>1\right)$
I took $\cos ^nx$ as $\cos ^{n-2}x\cos ^2x$ and replaced $\cos ^2x$ with $1-\sin ^2x$. Then integrated using by parts. Could not arrive at the given result
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Hint: use $\cos xdx = d\left(\sin x\right)$ : $$\eqalign{ & I_n = ∫_0^{{π \over 2}} {{{\cos }^n}x} dx = ∫_0^{{π \over 2}} {{{\cos }^{n - 1}}x} d\left(\sin x\right) \cr & = \left. {\sin x{{\cos }^{n - 1}}x} \right|_0^{{π \over 2}} + \left(n - 1\right)∫_0^{{π \over 2}} {{{\sin }^2}x{{\cos }^{n - 2}}x} dx \cr & = \left(n - 1\right)∫_0^{{π \over 2}} {\left(1 - {{\cos }^2}x\right){{\cos }^{n - 2}}x} dx \cr & = \left(n - 1\right){I_{n - 2}} - \left(n - 1\right){I_n} \cr & \Longrightarrow n{I_n} = \left(n - 1\right){I_{n - 2}} }$$
AsukaMinato
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Shine Mic
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I am not familiar with this type of substitution. I have never used it before – user140161 Oct 17 '14 at 15:42
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@user140161 $d(f(x)) = f'(x)dx$ – Shine Mic Oct 17 '14 at 15:44
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In the second line, did you use by parts? – user140161 Oct 17 '14 at 15:46
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@user140161 Of course. And note that $\left. {\sin x{{\cos }^{n - 1}}x} \right|_0^{{\pi \over 2}} = 0$ – Shine Mic Oct 17 '14 at 15:47
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@user140161 It is not a substitution. What shine did was, rewriting $cos^nx$ as $cos^{n-1}x$ times $cosx$ and on that product the formula of Integration by Parts is used. – imranfat Oct 17 '14 at 15:49