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.
 A: For all intents and purposes, this amounts to the integral
$$\int_0^{2\pi} \frac{1}{1+ a \cos x} \, \mathrm{d}x$$
for $a := \cos \theta$. This is a rational function of sine and cosine, and hence the Weierstrass substitution may apply, if be a bit messy. Then
$$t = \tan \frac x 2 \implies \cos x = \frac{1-t^2}{1+t^2} , \, \mathrm{d}x = \frac{2 \, \mathrm{d}t}{1+t^2}$$
The integral becomes, after splitting up at $\pi$,
$$ \int_0^\infty \frac{1}{1 + a \frac{1-t^2}{1+t^2}} \frac{2 \, \mathrm{d}t}{1+t^2}
+  \int_{-\infty}^0 \frac{1}{1 + a \frac{1-t^2}{1+t^2}} \frac{2 \, \mathrm{d}t}{1+t^2}$$
Recombining and simplifying, we get
$$2 \int_{-\infty}^\infty \frac{\mathrm{d}t}{1+t^2 + a(1-t^2)} = 2 \int_{-\infty}^\infty \frac{1}{t^2(1-a) + (1+a)} \, \mathrm{d}t$$
This can be solved using that
$$\int \frac{1}{u^2 + \beta^2} \, \mathrm{d}u = \frac 1 \beta \arctan \frac u \beta + C$$
A: Utilize the Fourier series below\begin{equation}
\frac{\sin\theta}{1+\cos\theta\cos x}=1+2\sum_{k=1}^\infty \left(\frac{\sin\theta-1}{\cos\theta}\right)^k\cos kx
\end{equation}
and recognize that only the constant term survives upon integration, i.e.
$$\int_0^{2 \pi} \frac{\sin\theta}{1+\cos \theta \cos x} d x=2\pi
$$
A: 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
A: Also using complex analysis for $\theta\ne \frac{\pi}{2}$ we have
$$
\int_0^{2 \pi} \frac{dz}{1+\cos \theta \cos x}
{=
\frac{1}{i}\oint_{|z|=1} \frac{2dz}{2z+\cos \theta (z^2+1)}
\\=
\frac{1}{i}2\pi i\text{Res}\left\{\frac{2}{2z+\cos \theta (z^2+1)}\right\}\Bigg|_{z=\frac{\sin \theta-1}{\cos\theta}}
\\=
2\pi\text{Res}\left\{\frac{2}{2z+\cos \theta (z^2+1)}\right\}\Bigg|_{z=\frac{\sin \theta-1}{\cos\theta}}
\\=
2\pi\left\{\frac{1}{1+z\cos \theta}\right\}\Bigg|_{z=\frac{\sin \theta-1}{\cos\theta}}
\\=
\frac{2\pi}{\sin \theta}.
}
$$
A: Thanks to @David G. Stork who gave an alternative method as below:
$$
\begin{aligned}
 I&= \int_0^{2 \pi} \frac{d x}{1+a \cos x} d x\\& =\int_0^{2 \pi} \frac{1-a \cos x}{1-a^2 \cos ^2 x} d x\\&=\int_0^{2 \pi} \frac{d x}{1-a^2 \cos ^2 x}-a \int_0^{2 \pi} \frac{\cos x}{1-a^2 \cos ^2 x} d x \\
& =4 \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{\sec ^2 x-a^2} d x-a \int_0^{2 \pi} \frac{d(\sin x)}{\left(1-a^2\right)+a^2 \sin ^2 x} \\
& =4 \int_0^{\frac{\pi}{2}} \frac{d(\tan x)}{\tan ^2 x+\left(1-a^2\right)}-\frac{1}{1-a^2}\left[\tan ^{-1}\left(\frac{a \sin x}{\sqrt{1-a^2}}\right)\right]_0^{2 \pi} \\
& =\frac{4}{\sqrt{1-a^2}}\left[\tan ^{-1}\left(\frac{\tan x}{\sqrt{1-a^2}}\right)\right]_0^{\frac{\pi}{2}} \\
& =\frac{2 \pi}{\sqrt{1-a^2}} \\
&
\end{aligned}
$$
Putting $a=\cos\theta$ yields our result.
A: Splitting the integral interval into two gives another solution.
$$
\begin{aligned}
\int_0^{2 \pi} \frac{1}{1+a \cos x} d x & =\int_0^\pi \frac{1}{1+a \cos x} d x+\int_\pi^{2 \pi} \frac{1}{1+a \cos x} d x \\
& =\int_0^\pi \frac{1}{1+a \cos x} d x+\int_0^\pi \frac{1}{1-a \cos x} d x \\
& =2 \int_0^\pi \frac{1}{1-a^2 \cos ^2 x} d x \\
& =4 \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{\sec ^2 x-a^2} d x \\
& =4 \int_0^{\frac{\pi}{2}} \frac{d(\tan x)}{\tan ^2 x+\left(1-a^2\right)} \\
& =\frac{4}{\sqrt{1-a^2}}\left[\tan ^{-1}\left(\frac{\tan x}{\sqrt{1-a^2}}\right)\right]_0^{\frac{\pi}{2}} \\
& =\frac{2 \pi}{\sqrt{1-a^2}}
\end{aligned}
$$
