TITLE: Find $\int^{\pi/2}_0 \frac{\sin^{n-2}(x)}{(1+\cos x)^n}\mathrm dx$
QUESTION [8 upvotes]: Finding 
  $$I=\int_0^{\pi/2}\frac{\sin^{n-2}(x)}{(1+\cos x)^n}\mathrm dx$$

What I tried: 
\begin{align*}
I &= \int^{\pi/2}_0 \left(\frac{\sin x}{1+\cos x}\right)^n\csc^2(x)\mathrm dx \\
  &= \int^{\pi/2}_0 \tan^n(x/2)\csc^2(x)\mathrm dx
\end{align*}
But I don't know how to proceed further. Could you please help?

REPLY [3 votes]: Sorry for the late answer, but I just saw this in my feed and I'd like propose one more way of finding the integral:
$$I=\int \frac{\sin^{n-2}(x)}{(1+\cos x)^n}dx$$
If you want to completely avoid trig identities:
Let $\cos x=u$ then $$I=-\int_1^0(1+u)^{-n}(1-u^2)^{(n-3)/2}du=-\int _1^0(1+u)^{-(n+3)/2}(1-u)^{(n-3)/2}du$$
Let $z=1-u$ then $$I=\int_0^1 (2-z)^{-(n+3)/2}z^{(n-3)/2}dz$$
Again: let $2-z=t^2z$ which leads to $$dt=-\frac{z^{-3/2}}{\sqrt{2-z}}dz$$ or $$-z^2tdt=dz$$
$$I=-\int _\infty^1 (2-z)^{-n/2 -1} z^{n}dt=-\frac 12 \int_\infty^1 t^{-n-2}+t^{-n }dt=\frac 12 \left(\frac{t^{-n-1}}{-n-1}+\frac{t^{-n+1}}{-n+1}\right)^\infty_1$$
And finally $$I=\frac 12 \left(\frac{1}{n-1}+\frac{1}{n+1}\right)=\frac n{n^2-1}$$