How do you convert this integral into a contour integral over the unit circle (such that you can use the residue theorem)? For context, this is to determine the coefficient $b_n$ in a Fourier series. $$\int_{0}^{2\pi} f(\theta)\text{sin}(n\theta) d\theta\ = \int_{0}^{2\pi} \frac{\text{sin}(\theta)}{5+4\text{cos}(\theta)}\text{sin}(n\theta) d\theta$$ where $$n=1,2,3,\ldots$$
How should I approach this? Namely, I know I can convert $\text{sin}(\theta)$ to $\frac{e^{i\theta}-e^{-i\theta}}{2i}=\frac{z-\frac{1}{z}}{2i}$ and similarly for $\text{cos}(\theta)$, but I am stumped by how to deal with the $\text{sin}(n\theta)$ and namely the $n$.
I have tried using the above approach such that $\text{sin}(n\theta)=\frac{e^{in\theta}-e^{-in\theta}}{2i}=\frac{z^n-\frac{1}{z^n}}{2i}$. However I am unsure how to find the residue after simplifying everything. After simplifying, we have poles at -2, -1/2, and 0 (order n+1), of which -1/2 and 0 are enclosed in the unit circle. Using the limit approach to find the residue for the pole at 0 with order n+1, we take the nth derivative and find the limit of that as z approaches 0: $$\text{Res}(f,0)=\frac{1}{n!}\lim_{z \to 0} \frac{d^n}{dz^n} \frac{(z-1)(z+1)(z^n-1)(z^n+1)}{(z+2)(z+\frac{1}{2})}$$
I am unsure how to do this part specifically. Is this the right approach? Did I mess something up?