Residue Theorem: Evaluate the integral $\int_\pi^{3\pi} \frac {dx}{5\cos x+13}$ Evaluate the integral using the Residue Theorem. $$\int_\pi^{3\pi} \frac {dx}{5\cos x+13}$$

Residue Theorem makes my head hurt. I have a lot of trouble with Laurent series in the first place. Any help would be greatly appreciated!
 A: The approach to evaluating this integral using contour integration is classical.  It begins with the substitution $z=e^{i x}$, which implies $dx=\frac{1}{iz}\,dz$.  The domain of integration transforms from $x\in [\pi,3\pi]$ to an integral on the unit circle $|z|=1$.
Proceeding as discussed, we can write
$$\begin{align}
\int_\pi^{3\pi}\frac{1}{5\cos(x)+13}\,dx&=\oint_{|z|=1}\left(\frac{1}{5\left(\frac{z+z^{-1}}{2}\right)+13}\right)\,\frac{1}{iz}\,dz\\\\
&=\frac2i \oint_{|z|=1}\frac{1}{5z^2+26z+5}\,dz\\\\
&=\frac2i \oint_{|z|=1}\frac{1}{(5z+1)(z+5)}\,dz\\\\
&=2\pi i \left(\frac2i\right)\text{Res}\left(\frac{1}{(5z+1)(z+5)},z=-1/5\right)\\\\
&=\frac{\pi}{6}
\end{align}$$
A: By exploiting symmetry and the residue theorem such integral is pretty simple to tackle.
Symmetry first:
$$ \int_{\pi}^{3\pi}\frac{dx}{5\cos x+13}=\int_{-\pi}^{\pi}\frac{dx}{13+5\cos(x)} = 2\int_{0}^{\pi}\frac{dx}{13+5\cos x}\\ = 2\int_{0}^{\pi/2}\frac{26}{13^2-5^2\cos^2 x}\stackrel{x\mapsto \arctan t}{=} 54\int_{0}^{+\infty}\frac{dt}{13^2(1+t^2)-5^2} $$
and the problem boils down to computing:
$$ 26\int_{-\infty}^{+\infty}\frac{dt}{12^2+13^2 t^2} = 26\cdot(2\pi i)\cdot\text{Res}\left(\frac{1}{12^2 + 13^2 t^2},t=\frac{12}{13}i\right)$$
or:
$$ 26\cdot(2\pi i)\cdot\left(-\frac{i}{312}\right) = \frac{54 \pi}{312} = \color{red}{\frac{\pi}{6}}.$$
