# Complex Variable Theory Integration

Integrate $\int_0^ {2\pi} \frac{sin^2\theta} {2 - cos\theta} d\theta$

I used the substitutions sin($\theta$) = $\frac{ z - z^{-1}}{2i}$ and cos($\theta$) = $\frac{ z + z^{-1}}{2}$ and d$\theta$ = $\frac{1}{iz}dz$ transforming the integral into $\frac{-i}{2} \oint \frac{1}{z^2}\frac{(z-1)^2}{z^2-4z+1}dz$ leaving me at the point where I am stuck.

• Very unclear what the function is you're integrating. Also, what have you tried. – Moya Feb 16 '16 at 4:54
• See here. – Mhenni Benghorbal Feb 16 '16 at 5:59
• Do you know the residue theorem? Can you find the poles inside the unit circle and the residues there? – Robert Israel Feb 16 '16 at 16:11

The function has period $2\pi$, so $I=\int\limits_0^{2\pi}\frac{\sin^2x\, dx}{2-\cos x}=\int\limits_{-\pi}^{\pi}\frac{\sin^2x\, dx}{2-\cos x}$,
also it is even function, so $I=2\int\limits_{0}^{\pi}\frac{\sin^2x\, dx}{2-\cos x}$. Now take $t=\tan\frac{x}2$