Integration (1 Viewer)

scardizzle · May 26, 2010

This may be an easy questions but I'm just not seeing it. Any help is greatly appreciated, cheers.

http://i.imgur.com/cpA9K.jpg

Trebla · May 26, 2010

For convenience I used $x$ instead of $\theta$

$u_n=\int_{0}^{\frac{\pi}{2}}x\sin^nx\,dx \\\\= \int_{0}^{\frac{\pi}{2}}x\sin^{n-2}x\sin^2x\,dx\\\\=\int_{0}^{\frac{\pi}{2}}x\sin^{n-2}x\,dx-\int_{0}^{\frac{\pi}{2}}x\sin^{n-2}x\cos^2x\,dx\\\\=u_{n-2}-\int_{0}^{\frac{\pi}{2}}x\sin^{n-2}x\cos^2x\,dx\\\\u=x\cos x,\,du=\cos x -x\sin x\,dx\\dv=\sin^{n-2}x\cos x\,dx,v=\dfrac{1}{n-1}\sin^{n-1}x\\\\u_n=u_{n-2}-\dfrac{1}{n-1}\left [ x\cos x\sin^{n-1}x \right ]_{0}^{\frac{\pi}{2}}+\dfrac{1}{n-1}\int_{0}^{\frac{\pi}{2}}\sin^{n-1}x\cos x\,dx-\dfrac{1}{n-1}\int_{0}^{\frac{\pi}{2}}x\sin^nx\,dx\\\\u_n=u_{n-2}+\dfrac{1}{n-1}\left [ \dfrac{\sin^nx}{n}\right ]_{0}^{\frac{\pi}{2}}-\dfrac{1}{n-1}u_n\\\\\dfrac{n}{n-1}u_n=u_{n-2}+\dfrac{1}{n(n-1)}\\\\\therefore u_n=\dfrac{n-1}{n}u_{n-2}+\dfrac{1}{n^2}$

seanieg89 · May 26, 2010

alternatively break it up as:

$x \sin^n(x) = x \sin^{n-1}(x)\sin(x)=-x \sin^{n-1}(x)\frac{d}{dx}(\cos(x))$ , then only one application of IBP will be required. more or less the same working though.

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Integration (1 Viewer)

scardizzle

Salve!

Trebla

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seanieg89

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