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Math Help - another integral reduction

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    another integral reduction

    see attachment for the question and what i did:
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    Quote Originally Posted by Pulock2009 View Post
    see attachment for the question and what i did:
    Integration by parts. Use an identifty with the \cos^2x and then add and subtract to get pieces like the denominator.
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    hey i didnot get u!could u show me some of the steps??Like i dunno what identity to use and therefore i dunno wat to add and subtract???
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    Quote Originally Posted by Pulock2009 View Post
    hey i didnot get u!could u show me some of the steps??Like i dunno what identity to use and therefore i dunno wat to add and subtract???
    This is what I was thinking but it doesn't really look like a reduction formula. Your integral, by parts

     <br />
\int \frac{\sin x}{(a+b \sin x)^{n+1}} dx = - \frac{\cos x}{(a+b \sin x)^{n+1}} - (n+1) b \int \frac{\cos^2 x}{(a+b \sin x)^{n+2}}dx<br />

     <br />
= - \frac{\cos x}{(a+b \sin x)^{n+1}} - (n+1) b \int \frac{1 - \sin^2 x}{(a+b \sin x)^{n+2}}dx<br />

     <br />
= - \frac{\cos x}{(a+b \sin x)^{n+1}} - (n+1) b I_{n+2} + (n+1) b\int \frac{\sin^2 x}{(a+b \sin x)^{n+2}}dx<br />

     <br />
= - \frac{\cos x}{(a+b \sin x)^{n+1}} - (n+1) b I_{n+2} + \frac{(n+1) }{b}\int \frac{b^2\sin^2 x}{(a+b \sin x)^{n+2}}dx<br />

     <br />
= - \frac{\cos x}{(a+b \sin x)^{n+1}} - (n+1) b I_{n+2} + \frac{(n+1) }{b}\int \frac{(a + b \sin x - a)^2 }{(a+b \sin x)^{n+2}}dx<br />

     <br />
= - \frac{\cos x}{(a+b \sin x)^{n+1}} - (n+1) b I_{n+2} + \frac{(n+1) }{b}\int \frac{(a + b \sin x)^2 - 2a(a + b \sin x ) + a^2 }{(a+b \sin x)^{n+2}}dx<br />

     <br />
= - \frac{\cos x}{(a+b \sin x)^{n+1}} - (n+1) b I_{n+2} + \frac{(n+1) }{b}(I_n - 2aI_{n+1} + a^2 I_{n+2}).<br />
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    great but why doesnot it look like a reduction formulae???
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    Quote Originally Posted by Pulock2009 View Post
    great but why doesnot it look like a reduction formulae???
    Well, I guess. You could solve this for I_{n+2}.
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