<s>[INST] 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!
 [/INST] 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:
$$ 26\cdot(2\pi i)\cdot\left(-\frac{i}{312}\right) = \frac{54 \pi}{312} = \color{red}{\frac{\pi}{6}}.$$
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