<s>[INST] Evaluating $\int_0^{2 \pi} \frac{1}{5-3 \sin z}dz$ So I need to evaluate $\int_0^{2 \pi} \frac{1}{5-3 \sin z}dz$. Here's what I have thus far:
\begin{align*}
\int_0^{2 \pi} \frac{1}{5-3 \sin z}dz&= \frac{1}{3} \int_0^{2 \pi} \frac{1}{\frac{5}{3}- \sin z}dz\\
&=\frac{1}{3} \int_0^{2 \pi} \frac{1}{\frac{10i-3e^{i z}+3e^{-i z}}{6i}} \\
&=2 i \int_0^{2 \pi} \frac{1}{10i - 3e^{iz}+3e^{-iz}}dz\\
&= 2 i \int_0^{2 \pi} \frac{1}{10i - 3e^{iz}+3e^{-iz}} \frac{e^{iz}}{e^{iz}}dz\\
&=2 \int_0^{2 \pi} \frac{1}{10iu-3u^2+3} du&& \text{where $u=e^{iz}$, $du=ie^{iz}dz$}\\
&= -2 \int_0^{2 \pi} \frac{1}{3(u+3i)(u+\frac{i}{3})} du
\end{align*}
but I get stuck here, Am I on the right path? Also I don't think that
$$-3(u+3i)(u+\frac{i}{3})=10iu-3u^2+3$$
Im off by a negative for the $10iu$. If possible, when helping use the formula for $\sin z$:
$$\frac{e^{iz}-e^{-iz}}{2i}.$$
Thanks in advance. (PS- why would someone downvote this? Its clearly showing I have work shown and have made an attempt at a solution!)
 [/INST] Let $t=\tan(\frac{z}{2})$, $\sin z=\frac{2t}{t^2+1}$ and $dz=\frac{2dt}{t^2+1}$ and thinking the integral from $-\pi$ to $\pi$, we get
$$\int_{-\infty}^{\infty}\frac{2}{5}\frac{1}{(t-\frac{3}{5})^2+(\frac{4}{5})^2}dt=\left.\frac{1}{2}\arctan\left(\frac{5t-3}{4}\right)\right\rvert_{-\infty}^{\infty}=\frac{\pi}{2}$$
</s>