Calculation of $\int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x}\mathrm dx$ 
Calculate the definite integral
$$
I=\int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x}\;\mathrm dx
$$
given that $a>b>0$.

My Attempt:
If we replace $x$ by $C$, then
$$
I = \int_{0}^{\pi}\frac{\sin^2 C}{a^2+b^2-2ab\cos C}\;\mathrm dC
$$
Now we can use the Cosine Formula ($A+B+C=\pi$). Applying the formula gives
$$
\begin{align}
\cos C &= \frac{a^2+b^2-c^2}{2ab}\\
a^2+b^2-2ab\cos C &= c^2
\end{align}
$$
From here we can use the formula $\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}$ to transform the integral to
$$
\begin{align}
I &= \int_{0}^{\pi}\frac{\sin^2 C}{c^2}\;\mathrm dC\\
&= \int_{0}^{\pi}\frac{\sin^2A}{a^2}\;\mathrm dC\\
&= \int_{0}^{\pi}\frac{\sin^2 B}{b^2}\;\mathrm dC
\end{align}
$$
Is my process right? If not, how can I calculate the above integral?

Such integral is related with the topological degree of a closed curve. We have:
$$ I = \frac{1}{2}\int_{0}^{2\pi}\frac{\sin^2 x}{\|a-be^{ix}\|^2}\,\mathrm dx = -\frac{i}{2}\oint_{\|z\|=1}\frac{\left(\frac{z-z^{-1}}{2}\right)^2}{(a-bz)(az-b)}\,\mathrm dz$$
hence we just have to consider the residues of the last integrand function in $z=0$ and $z=\dfrac{b}{a}$ (since $a>b>0$). That just leads to:
$$ I = \frac{\pi}{2a^2}.$$
