Integration - Reduction Method: PLEASE HELP!!!

**xwrathbringerx** · February 9th 2010, 07:01 PM

I've attempted both questions but I cant seem to get the answers. Could someone please show me where I got it wrong.

P.S. For the question that's typed up, I'm trying to find out the last part of the question i.e. "evaluate ..."

**dedust** · February 9th 2010, 07:21 PM

Originally Posted by xwrathbringerx

I've attempted both questions but I cant seem to get the answers. Could someone please show me where I got it wrong.

P.S. For the question that's typed up, I'm trying to find out the last part of the question i.e. "evaluate ..."

$I_{n+1} = \int \frac{\cos^{2(n+1)} x}{\sin x} ~dx = \cos^2 x \int \frac{\cos^{2n} x}{\sin x} ~dx = I_n \cos^2 x = I_n - I_n \sin^2 x$

$I_n \sin^2 x = \int \frac{\sin^2 x\cos^{2n} x}{\sin x} ~dx = \int \sin x \cos^{2n} x ~dx = - \frac{\cos^{2n+1} x}{2n + 1}$

hence,

$I_{n+1} = I_n - I_n \sin^2 x = I_n + \frac{\cos^{2n+1} x}{2n + 1}$

and the result follow

**xwrathbringerx** · February 9th 2010, 07:38 PM

I cant seem to be able to get the next part at all...

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