calculus - 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?

Answer

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
$$\begin{align} \int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x}\,\mathrm dx&=\frac{1}{2}\int_0^{\pi} \frac{1-\cos2 x}{a^2+b^2-2ab \cos x}\,\mathrm dx\\ &=\frac{1}{2}\left[\frac{\pi}{a^2-b^2}-\frac{\pi}{a^2-b^2}\left(\frac{b}{a}\right)^2\right]\\ &=\frac{\pi}{2 a^2} \end{align}$$