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?
 A: Using the formula $$a^2+b^2-ab\cos C = c^2\;,$$ Now If $C\leftrightarrow x\;,$ Then $a^2+b^2-2ab\cos x= c^2$
So Integral Convert into $$\displaystyle I = \int_{0}^{\pi}\frac{\sin^2 C}{c^2}\mathrm dC\;,$$ Now Using Sin formula $\displaystyle \frac{\sin C}{c} = \frac{\sin A}{a}.$
So $$\displaystyle I = \int_{0}^{\pi}\frac{\sin^2 A}{a^2}\mathrm dC\;,$$ Now $A+B+C = \pi\;,$ Then $\mathrm dC = 0-\mathrm dA-\mathrm dB$
Now when $C\rightarrow 0,$ Then $A\rightarrow \pi$ and $B\rightarrow 0$
When $C\rightarrow 0,$ Then $A\rightarrow 0$ and $B\rightarrow 0$
So $$\displaystyle I = \int_{\pi}^{0}\frac{\sin^2 A}{a^2}(-\mathrm dA)+\int_{0}^{0}\frac{\sin^2 A}{a^2}(-\mathrm dB)$$
So $$\displaystyle I = \int_{0}^{\pi}\frac{\sin^2 A}{a^2}\mathrm dA = \frac{2}{2a^2}\int_{0}^{\frac{\pi}{2}}\left[1-\cos 2A\right]\mathrm dA = \frac{\pi}{2a^2}$$
which agrees with the other solutions.
A: Let $p=a^2+b^2,q=2ab$ so that $p^2-q^2=(a^2-b^2)^2$ and the integrand can be rewritten as $$\frac{\sin^2x}{p-q\cos x} = \frac{1}{q}\left(\cos x+\frac{p}{q}+\frac{1}{q}\cdot\frac{q^2-p^2}{p-q\cos x} \right) $$ and thus the desired integral is equal to $$0+\frac{p\pi}{q^2}+\frac{q^2-p^2}{q^2}\int_{0}^{\pi}\frac{dx}{p-q\cos x} $$ The integral above is famously equal to $\pi/\sqrt{p^2-q^2}$ and hence the value of the integral in question is equal to $$\frac{\pi} {q^2}\cdot(p-\sqrt{p^2-q^2})=\frac{\pi}{2a^2}$$
A: We have
$$\eqalign{I_m(a,b)=
\int_0^\pi\frac{\cos(mx)}{a^2-2ab\cos x+b^2}\,\mathrm dx
&=\left\{\matrix{\frac{\pi}{a^2-b^2}&\hbox{if}&m=0\\\cr
\frac{\pi}{a^2-b^2}\left(\frac{b}{a}\right)^m&\hbox{if}&m\ne0
}\right.}$$
Proof can be seen here. Hence
\begin{align}
\int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x}\,\mathrm dx&=\frac{1}{2}\int_0^{\pi} \frac{1-\cos2 x}{a^2+b^2-2ab \cos x}\,\mathrm dx\\
&=\frac{1}{2}\left[\frac{\pi}{a^2-b^2}-\frac{\pi}{a^2-b^2}\left(\frac{b}{a}\right)^2\right]\\
&=\frac{\pi}{2 a^2}
\end{align}
A: 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:
$$ I = \frac{\pi}{2a^2}.$$
