How can I solve this integral using residue theorom?

Question

$$ \int_0^{2 \pi} \frac{\sin ⁡x}{(1+\frac{1}{2}\cos x)} dx $$ I want to solve this.

I have solved this here in the image but I have not reached a real number

Answer 1

Recall $\sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i}$ and $\cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}$. Use the substitution $z= e^{i\theta} $ then, $dz = i e^{i\theta} d\theta$.

The integral becomes:

$$ \frac{2}i\oint_C \frac{(z-1/z)dz}{4+z+1/z} = \frac{2}i\oint_C \frac{(z^2 -1)dz}{z^2 + 4z +1} $$ where $C$ is the unit circle around the origin. The poles of the denominator appear to be at $z_{\pm} = -2 \pm \sqrt{3} $. Hence, we see that our contour includes only $z_{+}$.

$Res(z_{+})=-2 + \sqrt{3}$. This implies $I=4 \pi (-2 + \sqrt{3})$. We need the imaginer part of this result since the initial integral was containing $sin(x)=Im[e^{i \theta}].$ Finally,

$\int_0^{2 \pi} \frac{\sin ⁡x}{(1+\frac{1}{2}\cos x)} dx =0.$