I am looking at the solution to an integral of trig functions. When I attempted I started with the same method of replacing the cosines with their complex form. However, I used the full form of cosine for the numerator as it is used in the denominator. The question is to prove that it is equal to pi/12. Which, the solutions do get (using residues). Wolfram Alpha also affirms this numerically.
These are the first two steps of the solutions. After this $z=e^{it}$ is substituted in and the singularities are found in the domain $\vert{z}\vert=1$ and the integral is found using Residue theory. All steps after the first make sense. $$ \int_0^{2\pi}\frac{\cos(3t)}{5-4\cos(t)}dt=\int_0^{2\pi}\frac{{e}^{3it}}{5-2({e}^{it}+{e}^{-it})}dt$$ Thoughts on why only the first term of cosine's complex form is used?
You can consider the real part of the RHS, i.e. $$\int_0^{2\pi}\frac{\cos(3t)}{5-4\cos(t)}dt=\operatorname{Re}\left(\int_0^{2\pi}\frac{{e}^{3it}}{5-2({e}^{it}+{e}^{-it})}dt\right)$$ and then discover that the imaginary part of that complex integral on the right is actually zero.
Note that by letting $z=e^{it}$ then $dz=izdt$. Therefore $$ \begin{align*} \int_0^{2\pi}\frac{e^{3it}}{5-4\cos(t)}dt &= \int_{|z|=1}\frac{z^3}{5-2(z+1/z)}\frac{dz}{iz}\\ &=-\frac{1}{2i}\int_{|z|=1}\frac{z^3}{(z-1/2)(z-2)}dz \\ &=-\pi\operatorname{Res}\left(\frac{z^3}{(z-1/2)(z-2)},z=1/2\right)=\\ &=-\pi\frac{(1/2)^3}{(1/2-2)}=\frac{\pi}{12}\in\mathbb{R}. \end{align*} $$