How to evaluate $\frac{1}{2\pi}\int \frac{1}{(1-\alpha \cos \omega)^2 +\alpha^2\sin^2 \omega } \, d\omega$

Question

May I please get help to solve the following expression, $$\frac 1 {2\pi}\int \frac 1 {(1-\alpha \cos \omega)^2 +\alpha^2\sin^2 \omega} \, d\omega$$

I have tried several ways, but couldn't get through. Thanks in advance.

Answer 1

Simply put $\tan\frac{\omega}{2}=t $ to get

$$\frac{1}{2\pi}\int\frac{dt}{(a^2+1)(t^2+1)-2a(1-t^2)}$$

The answer is therefore

$$\frac{1}{2\pi|a^2-1|}\arctan\left( \frac{|a+1|t}{|a-1|}\right)+c$$