Question

by contour techniques $\int$ 1/(2+cos$\theta$ ) limit 0 to 2$\pi$

Answer 1

$\int\limits_{0}^{2\pi} \frac{d\theta}{2+\cos \theta}=\int\limits_{0}^{2\pi} \frac{2d\theta}{4+e^{i\theta}+e^{-i\theta}}=\int\limits_{0}^{2\pi} \frac{2e^{i\theta}d\theta}{4e^{i\theta}+e^{2i\theta}+1}=$

$=\frac{2}{i}\int\limits_{0}^{2\pi} \frac{de^{i\theta}}{e^{2i\theta}+4e^{i\theta}+1}=\frac{2}{i}\oint\limits_{|z|=1}\frac{dz}{z^2+4z+1}=\frac{2}{i}\oint\limits_{|z|=1}\frac{dz}{(z+2+\sqrt{3})(z+2-\sqrt{3})}$

Only one pole, $z=-2+\sqrt{3}$, is inside the contour $\{|z|=1\}$, therefore, by the residue theorem we have

$\int\limits_{0}^{2\pi} \frac{d\theta}{2+\cos \theta}=\frac{2}{i}2\pi i\,{\rm res}_{z=-2+\sqrt{3}}\frac{1}{(z+2+\sqrt{3})(z+2-\sqrt{3})}=$

$=4\pi\lim\limits_{z\to -2+\sqrt{3}}\frac{1}{z+2+\sqrt{3}}=\frac{2\pi}{\sqrt{3}}$

Answer. $\frac{2\pi}{\sqrt{3}}$