1. ## another integral reduction

see attachment for the question and what i did:

2. Originally Posted by Pulock2009
see attachment for the question and what i did:
Integration by parts. Use an identifty with the $\displaystyle \cos^2x$ and then add and subtract to get pieces like the denominator.

3. 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???

4. Originally Posted by Pulock2009
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

$\displaystyle \int \frac{\sin x}{(a+b \sin x)^{n+1}} dx = - \frac{\cos x}{(a+b \sin x)^{n+1}} - (n+1) b \int \frac{\cos^2 x}{(a+b \sin x)^{n+2}}dx$

$\displaystyle = - \frac{\cos x}{(a+b \sin x)^{n+1}} - (n+1) b \int \frac{1 - \sin^2 x}{(a+b \sin x)^{n+2}}dx$

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

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

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

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

$\displaystyle = - \frac{\cos x}{(a+b \sin x)^{n+1}} - (n+1) b I_{n+2} + \frac{(n+1) }{b}(I_n - 2aI_{n+1} + a^2 I_{n+2}).$

5. great but why doesnot it look like a reduction formulae???

6. Originally Posted by Pulock2009
great but why doesnot it look like a reduction formulae???
Well, I guess. You could solve this for I_{n+2}.