Evaluate $\int_0^{\pi} \frac{\sin^2 \theta}{(1-2a\cos\theta+a^2)(1-2b\cos\theta+b^2)}\mathrm{d\theta}, \space 0Evaluate by complex methods
$$\int_0^{\pi} \frac{\sin^2 \theta}{(1-2a\cos\theta+a^2)(1-2b\cos\theta+b^2)}\mathrm{d\theta}, \space 0<a<b<1$$
Sis.

Using the series presented by Cortizol, the integral can be written as:
$$\int_0^{\pi} dx\left(\sum_{n=1}^{\infty} a^{n-1}\sin (nx)\right)\left(\sum_{m=1}^{\infty} b^{m-1}\sin (mx)\right)=\frac{1}{ab}\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} a^nb^m\int_0^{\pi} \sin(nx)\sin(mx)\,dx $$
Notice that for $n\ne m$, the integral is always zero, hence we look at only those cases where $n=m$ i.e
$$\frac{1}{ab}\sum_{n=1}^{\infty} (ab)^n\int_0^{\pi} \sin^2(nx)\,dx=\frac{\pi}{2ab}\sum_{n=1}^{\infty} (ab)^n=\boxed{\dfrac{\pi}{2}\dfrac{1}{1-ab}}$$
$\blacksquare$
