I am trying to calculate the following integral using complex transformation;
$$\int_0^{2\pi}\frac{\cos^3\theta}{1-2a \cos\theta+a^2}d\theta$$ where $$\left\lvert a \right\rvert<1$$ I am comfortable with sine or cosine with their 1st power, but 3rd power gave me a really hard time. Any ideas?
Notice that $1-2a\cos\theta+a^2 = (1-ae^{i\theta})(1-ae^{-i\theta})$, hence
$$\frac{1}{1-2a\cos\theta+a^2}=\sum_{j\geq 0}a^j e^{ji\theta}\sum_{k\geq 0}a^k e^{-ki\theta}\tag{1}$$ has a simple Fourier cosine series. The same holds for $\cos^3\theta$: $$ \cos^3\theta = \frac{3}{4}\cos(x)+\frac{1}{4}\cos(3x) \tag{2}$$ and since $\int_{0}^{2\pi}\cos(n\theta)\cos(m\theta)\,d\theta=\pi\,\delta(m,n) $, $$ \int_{0}^{2\pi}\frac{\cos^3\theta}{1-2a\cos\theta+a^2}\,d\theta=\pi\left[\frac{3}{4}\sum_{|j-k|=1}a^{j+k}+\frac{1}{4}\sum_{|j-k|=3}a^{j+k}\right]\tag{3}$$ that simplifies to: $$ \pi\left[\frac{3}{4}\sum_{j\geq 0}a^{2j+1}+\frac{3}{4}\sum_{j\geq 1}a^{2j-1}+\frac{1}{4}\sum_{j\geq 0}a^{2j+3}+\frac{1}{4}\sum_{j\geq 3}a^{2j-3}\right]$$ then to: $$\boxed{ \int_{0}^{2\pi}\frac{\cos^3\theta}{1-2a\cos\theta+a^2}\,d\theta = \color{red}{\frac{a (3+a^2)}{1-a^2}\cdot\frac{\pi}{2}}}\tag{4}$$ This question is strictly related with the residue theorem and the Poisson kernel.
Mr. Jack D'Aurizio, S.I. Hayek's book covers this topic and suggests using the residue theorem, by transforming the f(x) function to f(z) in complex plane.
$$z=e^{i\theta}$$ $$ sin\theta=\frac{1}{2i}(z-\frac{1}{z}) $$ $$ cos\theta=\frac{1}{2 }(z+\frac{1}{z}) $$ $$ d\theta=-i\frac{dz}{z} $$ and the integral $$I=\int_0^{2\pi}F(sin\theta,cos\theta)d\theta$$ becomes $$I=\int_C{f(z)dz}$$ on a unit circle which is centered at the origin. Still, I am not sure how to transform $$cos(3\theta)$$ with this method.