# Calculation of $\int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x} dx\;,$

Calculate the definite integral

$$I=\int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x}\;dx$$

given that $a>b>0$

My Attempt:

If we replace $x$ by $C$, then

$$I = \int_{0}^{\pi}\frac{\sin^2 C}{a^2+b^2-2ab\cos C}\;dC$$

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 $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$ to transform the integral to

\begin{align} I &= \int_{0}^{\pi}\frac{\sin^2 C}{c^2}\;dC\\ &= \int_{0}^{\pi}\frac{\sin^2A}{a^2}\;dC\\ &= \int_{0}^{\pi}\frac{\sin^2 B}{b^2}\;dC \end{align}

Is my process right? If not, how can I calculate the above integral?

#### Solutions Collecting From Web of "Calculation of $\int_0^{\pi} \frac{\sin^2 x}{a^2+b^2-2ab \cos x} dx\;,$"

We have
\eqalign{I_m(a,b)= \int_0^\pi\frac{\cos(mx)}{a^2-2ab\cos x+b^2}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}\, dx&=\frac{1}{2}\int_0^{\pi} \frac{1-\cos2 x}{a^2+b^2-2ab \cos x}\, 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}

Such integral is related with the topological degree of a closed curve. We have:

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

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 $$\displaystyle I = \int_{0}^{\pi}\frac{\sin^2 C}{c^2}dC\;,$$ Now Using Sin formula $\displaystyle \frac{\sin C}{c} = \frac{\sin A}{a}.$

So $$\displaystyle I = \int_{0}^{\pi}\frac{\sin^2 A}{a^2}dC\;,$$ Now $A+B+C = \pi\;,$ Then $dC = 0-dA-dB$

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 $$\displaystyle I = \int_{\pi}^{0}\frac{\sin^2 A}{a^2}(-dA)+\int_{0}^{0}\frac{\sin^2 A}{a^2}(-dB)$$

So $$\displaystyle I = \int_{0}^{\pi}\frac{\sin^2 A}{a^2}dA = \frac{2}{2a^2}\int_{0}^{\frac{\pi}{2}}\left[1-\cos 2A\right]dA = \frac{\pi}{2a^2}$$

Can anyone explain me is my Solution is Right or not,

Thanks