If $0<b<a,{\displaystyle \underset{0}{\overset{2o}{\int }}\frac{{\sin }^2\theta d\theta }{a-b\cos \theta }=}$

$I={\displaystyle \underset{0}{\overset{2\pi }{\int }}\frac{{\sin }^2\theta d\theta }{a-b\cos \theta }=2{\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{{\sin }^2\theta }{a-b\cos \theta }d\theta }}$ ……………….(i)

   $\underset{0}{\overset{\pi }{\int }}f\left(x\right)dx=\underset{0}{\overset{\pi }{\int }}f(\mathrm{\pi }-\mathrm{x})\mathrm{dx}$ then

                               =$2\underset{0}{\overset{\pi }{\int }}\frac{{\sin }^2\theta }{a+b\cos \theta }d\theta$ ……………..(ii)

   add eq (i) and (ii) then

                               $=2\underset{0}{\overset{\frac{\pi }{2}}{\int }}{\sin }^2\theta \left(\frac{1}{a-b\cos \theta }+\frac{1}{a+b\cos \theta }\right)d\theta$

                               $=4a{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\sin }^2\theta d\theta }{a^2-b^2{\cos }^2\theta }}$

                               $=\frac{4a}{b^2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(1-\frac{a^2-b^2}{a^2-b^2{\cos }^2\theta }\right)d\theta }$

                              $=\frac{4a}{b^2}\left[\frac{\pi }{2}-(a^2-b^2)\underset{0}{\overset{\infty }{\int }}\frac{{\sec }^2\theta d\theta }{a^2(1+{\tan }^2\theta )-b^2}\right]$

                              $=\frac{4a}{b^2}\left[\frac{\pi }{2}-(a^2-b^2)\underset{0}{\overset{\infty }{\int }}\frac{dt}{a^2{t}^2+a^2-b^2}\right]$

                             $=\frac{4a}{b^2}{\left[\frac{\pi }{2}-\frac{(a^2-b^2)}{a\sqrt{a^2-b^2}}{\tan }^{-1}\frac{at}{\sqrt{a^2-b^2}}\right]}_0^{\infty }$

                            $=\frac{2\pi }{b^2}\left[a-\sqrt{a^2-b^2}\right]$