Cauchy's integral with sin used in it

Question

$$ \int_{0}^{2 \pi} \frac{1}{5-3 \sin \theta} d \theta $$

hello, i think there are multiple ways to solve this question, but i need to use Cauchy's. would you help me with it? thanks

Answer 1

Let $z=e^{i\theta}$ and then $\sin\theta=\frac1{2i}(z-\frac1z), d\theta=\frac1{iz}dz$. So \begin{eqnarray} \int_{0}^{2 \pi} \frac{1}{5-3 \sin \theta} d \theta&=&\int_{|z|=1}\frac{1}{5-3\cdot\frac1{2i}(z-\frac1z)}\frac1{iz}dz\\ &=&\int_{|z|=1}\frac{2}{3z^2-10iz-3}dz\\ &=&\int_{|z|=1}\frac{2}{3(z-z_1)(z-z_2)}dz. \end{eqnarray} Note that $z_{1}=3i,z_2=\frac{i}{3}$ are roots of $3z^2-10iz-3=0$ but only $z_2$ is inside $|z|=1$ and hence $$ \int_{0}^{2 \pi} \frac{1}{5-3 \sin \theta} d \theta=2\pi i\frac{2}{3(z_2-z_1)}=\frac{\pi}{2}.$$