I have to compute $$\int_0^{2\pi}\frac{dx}{7+6\cos(x)}$$
First, I was going to find a contour, which is the unit circle. $z(\theta) = e^{i\theta}$, $0 \leq \theta \leq 2\pi$. Now, $\cos(\theta) = \frac{1}{2}(e^{i \theta} + e^{-i\theta}) = \frac{1}{2}(z + z^{-1})$; $d\theta = \frac{dz}{iz}$. I filled it in:
$$\oint_C \frac{dz}{iz(7+3(z+z^{-1}))} = \frac{1}{i}\oint_C\frac{dz}{7z + 3z^2 + 3} = \frac{1}{i}\oint_C\frac{dz}{(z-(-7/6 - \sqrt{13}/6))(z-(\sqrt{13}/6-7/6))}$$
The only pole in the contour is $z = \frac{\sqrt{13}}{6}-\frac{7}{6}$, so we will find the residue of this singularity:
Res$(f, \frac{\sqrt{13}}{6}-\frac{7}{6}) = \lim_{z \to \frac{\sqrt{13}}{6}-\frac{7}{6}} (z - (\frac{\sqrt{13}}{6}-\frac{7}{6})\frac{1}{(z-(-7/6 - \sqrt{13}/6))(z-(\sqrt{13}/6-7/6))} = \frac{3}{\sqrt{13}}$
So, $$\frac{1}{i}\oint_C\frac{dz}{(z-(-7/6 - \sqrt{13}/6))(z-(\sqrt{13}/6-7/6))} = 2\pi i \cdot \frac{3}{\sqrt{13}i} = \frac{6\pi}{\sqrt{13}}$$
But this is wrong, it has to be $\frac{2\pi}{\sqrt{13}}$.
Where did I make a mistake?