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 

Your integral is equal to $$2\int_{0}^{\pi}\frac{\sin^2x}{a-b\cos x} \, dx$$ and we can express the integrand as $$\frac{\cos x} {b} +\frac{a}{b^2}+\frac{b^2-a^2}{b^2(a-b\cos x)} $$ and hence the integral in question is $$2\cdot 0+\frac{2\pi a} {b^2}+\frac{2(b^2-a^2)}{b^2}\int_{0}^{\pi}\frac{dx}{a-b\cos x} $$ The last integral can be evaluated using a tricky substitution $$(a-b\cos x) (a+b\cos y) =a^2-b^2$$ to get $\pi/\sqrt{a^2-b^2}$ and hence the desired answer is $$\frac{2\pi a} {b^2}-\frac{2\pi\sqrt{a^2-b^2}}{b^2}=\frac{2\pi}{a+\sqrt{a^2-b^2}}$$
