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

Question

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

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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!

Answer 1

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

$$\begin{align} \int_\pi^{3\pi}\frac{1}{5\cos(x)+13}\,dx&=\oint_{|z|=1}\left(\frac{1}{5\left(\frac{z+z^{-1}}{2}\right)+13}\right)\,\frac{1}{iz}\,dz\\\\ &=\frac2i \oint_{|z|=1}\frac{1}{5z^2+26z+5}\,dz\\\\ &=\frac2i \oint_{|z|=1}\frac{1}{(5z+1)(z+5)}\,dz\\\\ &=2\pi i \left(\frac2i\right)\text{Res}\left(\frac{1}{(5z+1)(z+5)},z=-1/5\right)\\\\ &=\frac{\pi}{6} \end{align}$$

Answer 2

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