Question

By contour techniques$\smallint$ 1/(2+cos$\theta$) limit 0 to 2$\pi$ .

Answer 1

${\displaystyle\int}\dfrac{1}{\cos\left({\theta}\right)+2}\,\mathrm{d}{\theta}$

$={\displaystyle\int}\dfrac{\sec^2\left(\frac{{\theta}}{2}\right)}{\tan^2\left(\frac{{\theta}}{2}\right)+3}\,\mathrm{d}{\theta}$ ...simplify using trigonometric

Substitute 

$u=\dfrac{\tan\left(\frac{{\theta}}{2}\right)}{\sqrt{3}}$, $\dfrac{\mathrm{d}u}{\mathrm{d}{\theta}} = \dfrac{\sec^2\left(\frac{{\theta}}{2}\right)}{2\sqrt{3}}$

$\mathrm{d}{\theta}=\dfrac{2\sqrt{3}}{\sec^2\left(\frac{{\theta}}{2}\right)}\,\mathrm{d}u$

$={\displaystyle\int}\dfrac{2\sqrt{3}}{3u^2+3}\,\mathrm{d}u$

$={{\dfrac{2}{\sqrt{3}}}}{\displaystyle\int}\dfrac{1}{u^2+1}\,\mathrm{d}u$

$=\dfrac{2\arctan\left(u\right)}{\sqrt{3}}$ substitution u

$=[\dfrac{2\arctan\left(\frac{\tan\left(\frac{{\theta}}{2}\right)}{\sqrt{3}}\right)}{\sqrt{3}}]_0^{2\pi}$

$=\dfrac{2{\pi}}{\sqrt{3}}$