Evaluate $\int_{0}^{\pi} \frac{d\theta}{(2+\cos\theta)^2}$ How can one evaluate $\displaystyle\int_{0}^{\pi} \frac{d\theta}{(2+\cos\theta)^2}$?
My attempt:
$$\int_{0}^{\pi} \frac{d\theta}{(2+\cos\theta)^2} =  \frac{1}{2}\int_{0}^{2\pi} \frac{d\theta}{(2+\cos\theta)^2}$$
To find the singularity, I solve:
$ (2+\cos\theta)^2 = 0 $ and therefore, $\cos\theta =  -2$.
Substituting: $\cos z = \frac{e^{iz} + e^{-iz}}{2} = \frac{z + \frac{1}{z}}{2}$,
I find that $z = -2 + \sqrt{3} $ is the singular point that lies in the unit circle $|z| = 1$.
From this point, I have little idea how to go about solving this problem. I know I have to find the residue and then just sum them but to get the expression that would cancel out the pole is where I am currently stuck.

I will start from the point where you left off.  The integral may be written as, upon substitution of $z=e^{i \theta}$:
$$\begin{align}\frac{1}{2} \int_0^{2 \pi} \frac{d\theta}{(2+\cos{\theta})^2} &= -i 2 \oint_{|z|=1} dz \frac{z}{(z^2+4 z+1)^2}\\ &= -i 2 \oint_{|z|=1} dz \frac{z}{(z+2-\sqrt{3})^2 (z+2+\sqrt{3})^2}\end{align}$$
after a bit of algebra.  This integral is $i 2 \pi$ times the sum of the residues of the poles within the integration contour.  The only pole inside this contour, as you point out, is the pole at $z=-2+\sqrt{3}$.  The other pole at $z=-2-\sqrt{3}$ is outside this contour and is not counted.  
To compute the residue at this pole, note that we have double roots.  For such roots, we have to take a derivative:
$$\mathrm{Res}_{z=-2+\sqrt{3}} \frac{-i 2 z}{(z+2-\sqrt{3})^2 (z+2+\sqrt{3})^2} = \lim_{z \rightarrow -2+\sqrt{3}} \left [\frac{d}{dz} \frac{-i 2 z}{(z+2+\sqrt{3})^2} \right ]$$
I will leave the algebra to the reader; the result is $-i \sqrt{3}/9$.  The integral is then
$$\int_0^{\pi} \frac{d\theta}{(2+\cos{\theta})^2} = i 2 \pi \frac{-i \sqrt{3}}{9} = \frac{2 \sqrt{3}}{9} \pi$$
