prove that $$\int_{0}^{2\pi} \frac{1}{(a+cos(\theta))^2} d\theta = \frac{2\pi a}{(a^2-1)^\frac{3}{2}} \\\\ for\ a > 1 $$
My idea to solve this question was to find poles of function $\frac{1}{(a+cos(z))^2}$ and reduce the problem to calculating residues of this function but I don't know how to this(even finding poles doesn't seem easy)
Thanks.