Evaluating integral using residue

Question

Show that $\int_{0}^{2 \pi} \frac{d \theta}{(5-3\sin(\theta))^2}= \frac{5 \pi}{32}$.

I don't quite know what zi sm doing wrong.

I simplified the expression as follows $$\int_{0}^{2 \pi} \frac{d \theta}{(5-3\sin(\theta))^2}= \frac{1}{i} \int_{0}^{2 \pi} \frac{4z}{(3iz^{2}+10z-3i)^2}$$

From here I found the zeros to be $\frac{-1}{3i}, \frac{-3}{i}$.

I then went on to say that $\frac{-1}{3i}$ is our only pole.

Then when I plug it into the residue formula I get $$\lim_{z \rightarrow \frac{-1}{3i}}\frac{d}{dz}\left[\left(z-\left(\frac{-1}{3i}\right)\right)^{2} \frac{4z}{(3iz^{2}+10z-3i)^2}\right]$$.

And from here I am stuck. I am not sure if I messed up in my calculations or used the wrong pole or what? Can anyone point me in the right direction?

Answer 1

You are almost done.

Now factor the denominator of the last expression using: $$ (3i z^2+10z -3 i)=3i\left(z-\frac i3\right)(z-3 i) $$

Can you take it from here?

$$\begin{align}&\lim_{z \rightarrow \frac{i}{3}}\frac{d}{dz}\left[\left(z-\frac{i}{3}\right)^{2} \frac{4z}{(3iz^{2}+10z-3i)^2}\right]\\&=-\frac49\lim_{z \rightarrow \frac{i}{3}}\frac{d}{dz}\left[\frac{z}{(z-3i)^2}\right]\\&=-\frac49\lim_{z \rightarrow \frac{i}{3}}\left[\frac{-z-3i}{(z-3i)^3}\right]=\frac{5}{64}.\end{align}$$

A side note: the integral over $z$ is not from $0$ to $2\pi$. It is $\oint_{|z|=1}f(z)dz$.