I have been trying to solve the integral below using contour integration. $$\int_{0}^{\pi}\frac{\sin^2(x)}{a+\cos(x)}dx, \quad a>1.$$
I'd appreciate to see how you would solve it, since the fact that it goes to only $\pi$ and not to $2\pi$ gives me some complications when solving for the residues, after using the variable change $y=2x$.
The chances of me saying this were high: let us avoid contour integration.
By symmetry $$ \int_{0}^{\pi}\frac{\sin^2\theta}{a+\cos\theta}\,d\theta = \int_{0}^{\pi/2}\sin^2(\theta)\left[\frac{1}{a+\cos\theta}+\frac{1}{a-\cos\theta}\right]d\theta$$ equals $$ 2a\int_{0}^{\pi/2}\frac{\sin^2\theta}{a^2-\cos^2\theta}\,d\theta\stackrel{\theta\mapsto\arctan t}{=}2a\int_{0}^{+\infty}\frac{dt}{(1+t^2)(a^2+(a^2-1)t^2)} $$ or $$ \frac{\pi}{a+\sqrt{a^2-1}} $$ by partial fraction decomposition.