<s>[INST] Is there another simpler method to evaluate the integral $\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x , \textrm{ where } \theta \in (0, \pi)?$ Using ‘rationalization’, we can split the integral into two manageable integrals as:
$\displaystyle \begin{aligned}\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x = & \int_0^{2 \pi} \frac{1-\cos \theta \cos x}{1-\cos ^2 \theta \cos ^2 x} d x \\= & \int_0^{2 \pi} \frac{d x}{1-\cos ^2 \theta \cos ^2 x}-\cos \theta \int_0^{2 \pi} \frac{\cos x}{1-\cos ^2 \theta \cos ^2 x} d x \\= & 4 \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{\sec ^2 x-\cos ^2 \theta} d x+\int_0^{2 \pi} \frac{d(\cos \theta \sin x)}{\sin ^2 \theta+\cos ^2 \theta \sin ^2x} \\= & 4 \int_0^{\frac{\pi}{2}} \frac{d\left(\tan x\right)}{\sin ^2 \theta+\tan ^2 x}+\frac{1}{\sin \theta}\left[\tan ^{-1}\left(\frac{\cos ^2 \theta \sin x}{\sin \theta}\right)\right]_0^{2 \pi} \\= & \frac{4}{\sin \theta}\left[\tan ^{-1}\left(\frac{\tan x}{\sin \theta}\right)\right]_0^{\frac{\pi}{2}} \\= & \frac{4}{\sin \theta} \cdot \frac{\pi}{2} \\= & \frac{2 \pi}{\sin \theta}\end{aligned}\tag*{} $
Is there another simpler method to evaluate the integral?
Your comments and alternative methods are highly appreciated.
 [/INST] As a heuristic calclulation which can be proved later:
Let $I(a) = \int^{2\pi}_{0}\frac{dx}{1+a\cos(x)}$.
Using the geometric series:
\begin{align*}
\frac{1}{1+a\cos(x)} = \sum^{\infty}_{n=1} (-1)^na^n\cos^n(x)
\end{align*}
We have:
\begin{align*}
I(a) = \sum^{\infty}_{n=1} (-1)^na^n \int^{2\pi}_{0} \cos^n(x)dx
\end{align*}
Consider:
\begin{align*}
\int\cos^n(x)dx &=  \cos^{n-1}(x)\sin(x) + (n-1) \int\cos^{n-2}(x)\sin^2(x)dx
\end{align*}
This means $\int^{2\pi}_{0}\cos^n(x)dx = \frac{(n-1)}{n}\int^{2\pi}_{0 }\cos^{n-2}(x)dx$.  By iterating you would find the odd terms zero and the series expansion to be:
\begin{align*}
I(a) = (2\pi)\left(1 + \frac{1}{2}a^2 +  \frac{3}{8}a^4  ...\right)
\end{align*}
which would be the expansion of $2\pi(\sqrt{1-a^2})^{-1}$
I think we can justify using the uniform convergence of the geometric series when $|a|<1$ that we can switch summation and integration
</s>