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

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.

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

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

$$\frac{1}{2\pi|a^2-1|}\arctan\left( \frac{|a+1|t}{|a-1|}\right)+c$$
• First use that $\cos^2(x)+\sin^2(x)=1$. now when doing that substitution, you need to convert $\cos\omega$ in terms of t. That is given by $\cos(2u)=\frac{1-\tan^2(u)}{1+\tan^2(u)}$ . Feb 4, 2018 at 15:05
• After that use standard integral $\int \frac{dx}{a^2+x^2}$.. Feb 4, 2018 at 15:07