Computation of a certain integral I would like to compute the following integral. This is for a complex analysis course but I managed to around some other integrals using real analysis methodologies. Hopefully one might be able to do for this one too.
$$\int_{0}^{2\pi} \frac{1}{a-\cos(x)}dx, \text{ with } a > 1.$$
Any suggestion will be greatly appreciated.

$$\cos x = \frac{e^{i x}+e^{-ix}}{2}$$
Thus
$$\int_0^{2\pi} \frac{dx}{a-\frac{e^{i x}+e^{-ix}}{2}} = \oint_C \frac{1}{a-\frac{z+z^{-1}}{2}} \frac{dz}{iz} = - i\oint_C \frac{dz}{az-\frac{z^2+1}{2}} = 2i\oint_C \frac{dz}{z^2-2az+1}$$
where $C$ describes the unit circle $|z|=1$, centred at the origin, parametrized by $e^{iz}$ where $0\le z\le 2\pi$.
Letting $f(z)=\frac{1}{z^2-2az+1}$, we find that the poles of $f$ are at $z=a\pm\sqrt{a^2-1}$.  Noting that $a>1$, the only pole in $C$ is the one with the negative sign.  Then
$$\operatorname*{Res}_{z = a-\sqrt{a^2-1}}f(z)= \lim_{z\to a-\sqrt{a^2-1}}\frac{z-a+\sqrt{s^2+1}}{z^2-2az+1}= \lim_{z\to a-\sqrt{a^2-1}}\frac{1}{2z-2a}=- \frac{1}{2\sqrt{a^2-1}}$$
And thus we wrap up:
$$\int_0^{2\pi} \frac{dx}{a-\cos x} = 2i\oint_C \frac{dz}{z^2-2az+1} = 2i\left(-\frac{2\pi i }{2\sqrt{a^2-1}}\right) = \frac{2\pi}{\sqrt{a^2-1}}$$
$\blacksquare$
