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?
Best Answer
Note that the $\csc^2(x)$ term can be expressed in terms of $\tan\left(\frac x2\right)$ aswell. To be precise we got that
\begin{align*} \csc^2(x)=\left(\frac1{\sin(x)}\right)^2=\left(\frac1{2\sin\left(\frac x2\right)\cos\left(\frac x2\right)}\right)^2=\left(\frac{\frac1{\cos^2\left(\frac x2\right)}}{2\frac{\sin\left(\frac x2\right)}{\cos\left(\frac x2\right)}}\right)^2 =\left(\frac{1+\tan^2\left(\frac x2\right)}{2\tan\left(\frac x2\right)}\right)^2 \end{align*}
Using this and further noticing that $\frac{\mathrm d}{\mathrm dx}\tan\left(\frac x2\right)=\frac12\left(1+\tan^2\left(\frac x2\right)\right)$ we may enforce the substition $\tan\left(\frac x2\right)=u$ to obtain
\begin{align*} I_n=\int_0^{\pi/2}\tan^n\left(\frac x2\right)\csc^2(x)\mathrm dx&=\int_0^{\pi/2}\tan^n\left(\frac x2\right)\left(\frac{1+\tan^2\left(\frac x2\right)}{2\tan\left(\frac x2\right)}\right)^2\mathrm dx\\ &=\frac12\int_0^{\pi/2}\tan^{n-2}\left(\frac x2\right)\left(1+\tan^2\left(\frac x2\right)\right)\left[\frac12\left(1+\tan^2\left(\frac x2\right)\right)\mathrm dx\right]\\ &=\frac12\int_0^1 u^{n-2}(1+u^2)\mathrm du\\ &=\frac12\left[\frac{u^{n-1}}{n-1}+\frac{u^{n+1}}{n+1}\right]_0^1\\ &=\frac12\left[\frac1{n-1}+\frac1{n+1}\right] \end{align*}