If $0<b<a,{\int }_0^{2\pi }\frac{{\sin }^2\theta d\theta }{a-b\cos \theta }=$

$I={\int }_0^{2\pi }\frac{{\sin }^2\theta d\theta }{a-b\cos \theta }=2{\int }_0^{\pi }\frac{{\sin }^2\theta }{a-b\cos \theta }d\theta$

$=2{\int }_0^{\frac{\pi }{2}}{\sin }^2\theta \left(\frac{1}{a-b\cos \theta }+\frac{1}{a+b\cos \theta }\right)d\theta =4a{\int }_0^{\frac{\pi }{2}}\frac{{\sin }^2\theta d\theta }{a^2-b^2{\cos }^2\theta }$

$=\frac{4a}{b^2}{\int }_0^{\frac{\pi }{2}}\left(1-\frac{a^2-b^2}{a^2-b^2{\cos }^2\theta }\right)d\theta$

$=\frac{4a}{b^2}\left[\frac{\pi }{2}-\left(a^2-b^2\right){\int }_0^{\infty }\frac{{\sec }^2\theta d{\theta }^{\text{'}}}{a^2\left(1+{\tan }^2\theta \right)-b^2}\right]$

$=\frac{4a}{b^2}\left[\frac{\pi }{2}-\left(a^2-b^2\right){\int }_0^{\infty }\frac{dt}{a^2{t}^2+a^2-b^2}\right]$

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