see attachment for the question and what i did:
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}).
$