Do we have any residue formula to value the integral:
$$\int_{0}^{2 \pi}\frac{\sin^2 x}{ 5 + 4 \cos x} dx$$
I mean without finding out the pole and applying the residues theorem each time.
Any fast and convenient approach will be appreciated.
I have found a very useful formula which is $$\int_{0}^{\infty}\frac{dx}{1 +x^n}= \frac{\pi}{n} cosc \frac{\pi}{n}$$
I can evaluate a lot of integral using the theorem.
$$\begin{eqnarray*}\int_{0}^{2\pi}\frac{\sin^2(x)}{5+4\cos x}\,dx &=& \int_{0}^{\pi}\left(\frac{\sin^2(x)}{5+4\cos x}+\frac{\sin^2(x)}{5-4\cos x}\right)\,dx\\&=&10\int_{0}^{\pi}\frac{\sin^2(x)\,dx}{25-16\cos^2(x)}\\&=&20\int_{0}^{\pi/2}\frac{\sin^2(x)\,dx}{25-16\cos^2(x)}\\ (x=\arctan t)\quad &=&20\int_{0}^{+\infty}\frac{t^2}{(1+t^2)(9+25t^2)}\,dt\\&=&10\int_{-\infty}^{+\infty}\frac{t^2}{(1+t^2)(9+25t^2)}\,dt\\&=&20\pi i\sum_{\xi\in\{i,3i/5\}}\text{Res}\left(\frac{z^2}{(1+z^2)(9+25z^2)},z=\xi\right)=\color{red}{\frac{\pi}{4}}.\end{eqnarray*}$$
