How many methods are there to find $\int_{0}^{\pi} \frac{d \theta}{a+b \cos \theta}, \textrm{ where }a>b>0$ When I deal with this simple integral, I found there are several methods. Now I share one of them.
Letting $x\mapsto \pi-x $ yields $$I=\int_{0}^{\pi} \frac{d \theta}{a+b \cos \theta}=\int_{0}^{\pi} \frac{d \theta}{a-b \cos \theta}$$
Adding two versions together yields
$$
\begin{aligned}
2 I &=2 a \int_{0}^{\pi} \frac{d \theta}{a^{2}-b^{2} \cos ^{2} \theta} \\
&=4 a\int_0^{\frac{\pi}{2}} \frac{\sec ^{2} \theta}{a^{2} \sec ^{2} \theta-b^{2}} d \theta \quad \textrm{( By symmetry )}\\
&=4 a{\int_{0}^{\infty}} \frac{d t}{\left(a^{2}-b^{2}\right)+a^{2} t^{2}} \\
&=\frac{4}{\sqrt{a^{2}-b^{2}}}\left[\tan^{-1} \left(\frac{at}{\sqrt{a^{2}-b^{2}}}\right)\right]_{0}^{\infty} \\
&=\frac{2 \pi}{\sqrt{a^{2}-b^{2}}}
\end{aligned}
$$
We can now conclude that $$
\boxed{I=\frac{\pi}{\sqrt{a^{2}-b^{2}}}}
$$
Are there any other methods to deal with the integral?

By Weierstrass substitution
Letting $t=\tan \frac{\theta}{2}$ transforms the integral
$$
\begin{aligned}
I&=\int_{0}^{\pi} \frac{1}{a+\frac{b\left(1-t^{2}\right)}{1+t^{2}}} \frac{2 d t}{1+t^{2}}\\
&=2 \int_{0}^{\infty} \frac{d t}{(a+b)+(a-b) t^{2}}\\
&=\frac{2}{\sqrt{a-b} \sqrt{a+b}} \left[\tan ^{-1}\left(\frac{\sqrt{a-b}t}{\sqrt{a+b}}\right)\right]_{0}^{\infty}\\
&=\frac{\pi}{\sqrt{a^{2}-b^{2}}}
\end{aligned}
$$
