Finding value of $\int^{2\pi}_{0}\frac{\sin^2(x)}{a-b\cos x}dx$ without contour Integration 
Finding value of $\displaystyle \int^{2\pi}_{0}\frac{\sin^2(x)}{a-b\cos x}dx\;\;, a>b>0$ without contour Integration

Try: Let $\displaystyle I =\int^{2\pi}_{0}\frac{\sin^2 x}{a-b\cos x}dx=2\int^{\pi}_{0}\frac{\sin^2 x}{a-b\cos x}dx$
Using Integration by parts
$$ I =\frac{2}{b}\bigg[\sin (x)\cdot \ln|a-b\cos x|\bigg]\bigg|^{\pi}_{0}-\frac{2}{b}\int^{\pi}_{0}\cos x\cdot \ln|a-b\cos x|dx$$
$$I=-\frac{2}{b}\int^{\pi}_{0}\cos x\cdot \ln(a-b\cos x)dx$$
Could some help me to solve it, Thanks in advance 

From $a>b$ then $\left|\dfrac{b}{a}\cos x\right|<1$ with the expansion
\begin{align}
\int_{0}^{2\pi}\frac{\sin^2(x)}{a-b\cos x}\ dx 
&= \dfrac{1}{a}\int_{0}^{2\pi}\sin^2(x)\sum_{n\geq0}\left(\dfrac{b}{a}\cos x\right)^n\ dx \\
&= \dfrac{1}{a}\sum_{n\geq0}\left(\dfrac{b}{a}\right)^n\int_{0}^{2\pi}\sin^2x\cos^nx\ dx \\
&= \dfrac{1}{a}\sum_{n\geq0, \text{even}}\left(\dfrac{b}{a}\right)^n\left(\dfrac{2\pi}{2^n}{n\choose n/2} - \dfrac{2\pi}{2^{n+2}}{n+2\choose n/2+1}\right) \\
&= \color{blue}{\dfrac{2\pi}{a+\sqrt{a^2-b^2}}}
\end{align}
Using here
$$\int_0^{2\pi}\cos^n x\ dx=\dfrac{2\pi}{2^n}{n\choose n/2}$$
for even $n$ and for odd $n$ it is zero.
