<s>[INST] 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?
 [/INST] 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.
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