Within a problem with several steps, I am asked to show the following equality given that $\theta \in (0, \pi)$:
$$\pi\int _0^{\frac{\pi }{2}}\frac{1}{1+\cos \theta \cdot \cos x} \ \mathrm{d}x=\frac{\pi \theta}{\sin \theta}$$
I have no idea how to attack this. I don't see any clear trigonometric identity that I could apply. What would you suggest? Any hint/clue/help would be appreciated. Thanks.
You can check this post if you are interested in the rest of the problem.
$$\begin{eqnarray*}\int_{0}^{\pi/2}\frac{dx}{1+\cos(\theta)\cos(x)}&\stackrel{x\mapsto 2z}{=}&2\int_{0}^{\pi/4}\frac{dz}{1+\cos(\theta)(2\cos^2 z-1)} \\&\stackrel{z\mapsto\arctan t}{=}&2\int_{0}^{1}\frac{dt}{(1+t^2)+\cos(\theta)(1-t^2)}\end{eqnarray*}$$ is an elementary integral that equals $\frac{2}{\sin\theta}\arctan\left(\tan\frac{\theta}{2}\right)=\frac{\theta}{\sin\theta}$ as wanted.
Letting $t=\tan \frac{x}{2} $ yields $$ \begin{aligned} & \int_0^{\frac{\pi}{2}} \frac{1}{1+\cos \theta \cos x} d x \\ = & \int_0^1 \frac{1}{1+\cos \theta \frac{1-t^2}{1+t^2}} \cdot \frac{2 d t}{1+t^2} \\ = & 2 \int_0^1 \frac{1}{(1+\cos \theta)+(1-\cos \theta) t^2} d t \\ = & \frac{2}{\sqrt{(1+\cos \theta)(1-\cos \theta)}}\left[\tan \left(\sqrt{\frac{1-\cos \theta}{1+\cos \theta}} \cdot t\right)\right]_0^1 \\ = & \frac{2}{\sin \theta} \cdot \frac{\theta}{2} \\ = & \frac{\theta}{\sin \theta} \end{aligned} $$
