Using the Residue Theorem to Prove that $\int^{2\pi}_{0} \frac{1}{(a+\cos\theta)^{2}} d \theta=\frac{2\pi a}{(a^{2}-1)^{3/{2}}}.$ How do you evaluate the following integral? Here we take $a>1$.
$$\int^{2\pi}_{0} \frac{1}{(a+\cos\theta)^{2}} d \theta=\frac{2\pi a}{(a^{2}-1)^{\frac{3}{2}}}.$$
I know I have to use the Residue Theorem, however, I am stuck on which contour to use, and also how to find the pole of the function. Any hints are greatly appreciated.

Use the contour $|z| = 1$
First change the cosines into exponential forms.
$$\large\int \frac {1}{(a+\frac{e^{it}}{2} + \frac{e^{-it}}{2})^2} \ d\theta$$
$$z = e^{i\theta}\\ d\theta = \frac {1}{iz} dz$$
$$\large \oint_{|z| = 1} \frac {1}{iz(a+\frac{z}{2} + \frac{z^{-1}}{2})^2} \ dz$$
Which simplifies to:
$$\large \oint_{|z| = 1} \frac {4z}{i(z^2 + 2az+ 1)^2} \ dz\\
\oint_{|z| = 1} \frac {4z}{i(z+ a + \sqrt {a^2-1})^2(z+ a - \sqrt {a^2-1})^2} \ dz$$
Has one pole inside the contour.
The residual at $z = -a+\sqrt {a^2 - 1} = 2\pi i \frac {d}{dz} \frac {4z}{i(z+ a + \sqrt {a^2-1})^2}$ evaluated at $z =  -a+\sqrt {a^2 - 1}$
