integrating $ \frac{\cos(3\theta)}{5-3\cos(\theta)}d\theta$ using residue theorem

Question

So I've been taking complex analysis for fun recently and come across this problem: calculate $\int_{0}^{2\pi} \frac{\cos(3\theta)}{5-3\cos(\theta)} d\theta$ using the residue theorem, after making this a complex integral: $\int_{\partial B_+(0,1)}^{} \frac{z^3+z^{-3}}{-3iz^2+10iz-3i}$ (with $z = e^{i\theta}$) and after finding the singularities ($\frac{1}{3}$ and $3$) I then end up getting the as sum of the residues $-\frac{365i}{108}$ (is only the one for $\frac{1}{3}$ as the other one isn't in my ball $\partial B_+(0,1)$) and thus the answer I get is $\frac{365 \pi}{54}$ which is obviously wrong as Mark Fischler pointed out that the integral has a maximum of $\frac{1}{2}$ and thus can't exceed $\pi$ integral calculator is also saying that the answer should be $\frac{\pi}{54}$ where did I go wrong?

Answer 1

There's also an order-$3$ pole at $0$, which makes a contribution to the integral of$$\frac{2\pi i}{2!}\lim_{z\to 0}\frac{d^2}{dz^2}\frac{z^6+1}{-3iz^2+10iz-3i}=\pi\lim_{z\to 0}\frac{d^2}{dz^2}\frac{z^6+1}{-3z^2+10z-3}$$(see here). I'm lazy, so I'll compute the second-order derivative with Woflram Alpha, which gives the messy layout$$(z^6+1)\left(\frac{2(10-6z)^2}{(-3z^2+10z-3)^3}+\frac{6}{(-3z^2+10z-3)^2}\right)\\-\frac{12(10-6z)z^5}{(-3z^2+10z-3)^2}+\frac{30z^4}{-3z^2+10z-3}.$$At $z=0$, this is $\frac{200}{-27}+\frac69=-\frac{182}{27}$. So to your $\frac{365\pi}{54}$ we must add $-\frac{182\pi}{27}$, for a final result of $\frac{\pi}{54}$ as desired.