# Evaluate:$\int \limits_0^ {2\pi} \frac{\sin 2\theta}{a-b \cos \theta}d\theta$

Toolbox:
• $\int \limits _a^b f(x)dx=F(b)-F(a)$
• $\int \limits_0^{2a} f(x)dx=0$ when $f(2a-x)=-f(x)$
Given $I=\int \limits_0^ {2\pi} \large\frac{\sin 2\theta}{a-b \cos \theta}d\theta$
Let us replace f(x) as f(2a-x)
then $f(2 \pi-\theta)=\large\frac{\sin 2(2 \pi-\theta)}{a-b \cos (2 \pi-\theta)}=\frac{\sin (4 \pi-2 \theta)}{a-b \cos \theta}$
$\cos (2 \pi-\theta)=\cos \theta$
But $\sin (4 \pi-2 \theta)=-\sin 2 \theta$
Therefore $f(2 \pi-\theta)=\large\frac{-sin 2 \theta}{(a-b \cos \theta)}=-f(x)$
Hence by property $\int \limits_0^{2a} f(x)dx=0$ when $f(2a-x)=-f(x)$
Therefore $\int \limits_0^ {2\pi} \large\frac{\sin 2\theta}{a-b \cos \theta}d\theta=0$