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integral evaluation

Posted December 9th, 2007 by nico
in

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$$\int_0^{\frac{\pi}{2}}\sin^{2n}x\:dx=\frac{\pi}{2}\frac{1}{2}\frac{3}{4}\frac{5}{6}...\frac{2n-1}{2n}$$
$$\int_0^{\frac{\pi}{2}}\sin^{2n+1}x\:dx=\frac{2}{3}\frac{4}{5}\frac{6}{7}...\frac{2n}{2n+1}$$
‹ Steffensen's inequality proof Coordinate Geometry ›

Comments

Denote =>=> => => =>Same

On November 28th, 2011 MSA says:

Denote $ I_{2n}=\int^{\frac{\pi}{2}}_0 sin^{2n}x\,dx $ =>$ I_0=\int^{\frac{\pi}{2}}_0 1\,dx=\frac{\pi}{2} $
$ I_{2n}=\int^{\frac{\pi}{2}}_0 \sin^{2n-1}x(-\cosx)'\,dx=\sin^{2n-1}(\frac{\pi}{2})(-\cos\frac{\pi}{2})+\sin^{2n-1}0(\cos0)+\int^{\frac{\pi}{2}}_0(2n-1)\cos^2x\sin^{2n-2}x\,dx=(2n-1)\int^{\frac{\pi}{2}}_0(1-\sin^2x)\sin^{2n-2}x\,dx=<br /> (2n-1)\int^{\frac{\pi}{2}}_0\sin^{2n-2}x\,dx-(2n-1)\int^{\frac{\pi}{2}}_0 \sin^{2n}x\,dx=(2n-1)I_{2n-2}-(2n-1)I_{2n}} $

=> $ 2nI_{2n}=(2n-1)I_{2n-2} $=>$ I_{2n}=\frac{2n-1}{2n}I_{2n-2} $ =>$ I_{2n}=\frac{2n-1}{2n}...\frac{1}{2}I_0 $ =>$ I_{2n}=\frac{2n-1}{2n}...\frac{1}{2}\frac{\pi}{2} $

Same for $ I_{2n+1} $

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