Question

Using Residue theorem evaluate ∫0^2π dθ/2+sinθ ?

Answer 1

Make the substitution:$z = e^{ iθ} , dz = ie^{iθ}dθ.$ Now the complex $z$ describes the unit circle $C_1$ in the positive sense as $θ$ varies from to $2\pi.$

The integral becomes

$$\displaystyle\int_{0}^{2\pi}\dfrac{d\theta}{2+\sin \theta}=\oint_{C_1}\dfrac{\dfrac{dz}{iz}}{2+\dfrac{z^2-1}{2iz}}$$

$$=\oint_{C_1}\dfrac{2dz}{z^2+4iz-1}$$

The poles of the integrand are simple and occur when $z^2+4iz-1=0$

$$z_{\pm}=\dfrac{-4i\pm \sqrt{-16+4}}{2}=(-2\pm\sqrt{3})i$$

$$z_+z_-=(-2+\sqrt{3})i(-2-\sqrt{3})i=-1$$

Therefore, $z_+$ is a (simple) pole inside the unit circle.

Use $Res_{z=z_0}\dfrac{g(z)}{h(z)}=\dfrac{g(z_0)}{h'(z_0)},$ where $z_0$ is a simple zero of $h(z).$

Then

$$Res(z_+)=\dfrac{2}{2z_{+}+4i}=\dfrac{1}{-2i+\sqrt{3}i+2i}=\dfrac{1}{\sqrt{3}i}$$

Thus, by the residue theorem,

$$\displaystyle\int_{0}^{2\pi}\dfrac{d\theta}{2+\sin \theta}=2\pi i(\dfrac{1}{\sqrt{3}i})=\dfrac{2\pi}{\sqrt{3}}$$