Question

$\intop$ 1/(3-2cos$\theta$ +sin$\theta$ ) using contour integration limit 0 to 2pi

Answer 1

$1/(3-2cos\theta +sin\theta )=-\frac{2(1-tan^2(\theta /2))}{tan^2(\theta /2)+1}+\frac{2tan(\theta /2)}{tan^2(\theta /2)+1}+3$

$u=tan(\theta /2)\implies dx=2du/(u^2+1)$

$∫ 1/(3-2cos\theta +sin\theta)d\theta =2\int \frac{1}{2(u^2+u)+3u^2+1}du=2\int \frac{1}{(u\sqrt 5+1/\sqrt 5)^2+4/5}du=$

$v=(5u+1)/2\implies du 2dv/5$

$=2\int \frac{2}{5(4v^2/5+4/5)}dv=\int \frac{dv}{1+v^2}=arctanv=arctan \frac{5u+1}{2}=arctan\frac{5tan(\theta/2)+1}{2}$

$∫^{2\pi}_0 1/(3-2cos\theta +sin\theta)d\theta=arctan\frac{5tan(\theta/2)+1}{2}|^{2\pi}_0=$

$=arctan 1/2-arctan 1/2=0$