How to find the following definite integral:
$$\int_0^{\pi}\frac{\sin^{m-1}x}{(2+\cos x)^m}dx$$
I have done up to:
$$\int_0^{\pi}\frac{\sin^{m-1}x}{(2+\cos x)^m}dx=\int_0^{\pi}\frac{2^{m-1}\sin^{m-1}x/2~\cos^{m-1}x/2}{(3-2\sin^2x/2)^m}dx$$
Now, substituting $\sin^2x/2=z$, we have:
$$\int_0^1\frac{2^{m-1}z^{m/2-1}~(1-z)^{m/2-1}}{(3-2z)^m}dz$$
The numerator comes in the form of the beta integral... but what to do with the denominator ?
Help needed. Thank you !
The defined integral can be expressed thanks to the Incomplete Beta function (given by Lucian) or Gauss hypergeometric function. In case of integer $m$, an explicit form with Gamma function can be derived, which reduces to factorials if $m$ is even :