integration reduction (1 Viewer)

cutemouse · Jun 4, 2009

$If\ I_n = \int^\frac{\pi}{2}_0 sin^nxcos^2xdx\ for\ n\ge0\ show\ that\ I_{n-1}=\frac{n-1}{n+2}I_{n-2}\ for\ n\ge2$

Thanks

Dumbledore · Jun 4, 2009

start\: with \frac{1}{n+1}\int_{0}^{pi/2}[(n+1)sin^{n}x.cos\, x]cos\, x\, dx\\then \frac{1}{n+1}\int_{0}^{pi/2}cos\, x\, d(sin^{n+1}x) \: then \: by\: parts

wtf is this how do u get this to work?

gurmies · Jun 4, 2009

On behalf of Dumbledore:

$start\: with \frac{1}{n+1}\int_{0}^{pi/2}[(n+1)sin^{n}x.cos\, x]cos\, x\, dx\\then \frac{1}{n+1}\int_{0}^{pi/2}cos\, x\, d(sin^{n+1}x) \: then \: by\: parts$

integration reduction (1 Viewer)

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