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!

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:
$$ 26\cdot(2\pi i)\cdot\left(-\frac{i}{312}\right) = \frac{54 \pi}{312} = \color{red}{\frac{\pi}{6}}.$$
