Solve integral by finding Fourier series of complex function

I've mostly worked with real-valued functions, but this seems to be a complex-valued function and the integral for the coefficient doesn't seem that nice, especially when I rewrite the function as $$u(x)=\frac{r-e^{-ix}}{r^2-2r\cos x+1}.$$ I'm stuck on where to even start. Any ideas? I prefer to solve this via Fourier series if this is possible, and not using some other trick.

Try the function

\begin{align*}
v(x) = A + u(x) e^{ix}
\end{align*}

where ##A## is an appropriately chosen constant. You only need the real part of the function.

I solved it using Fourier series. It's pretty straightforward actually. The function is $$u(x)=\frac1{r-e^{ix}}=\frac1{r}\frac{1}{1-\frac{e^{ix}}{r}},$$ where ##r>1##. You see, this is the sum of a geometric series. Writing that out you will obtain the complex Fourier series and thus the coefficients. Then to solve the integral, you simply apply Parseval's formula. Bingo!

Yes, that is how you obtain the Fourier coefficients. I was thinking

\begin{align*}
c_0 = \frac{1}{2 \pi} \int_{-\pi}^\pi \dfrac{r - \cos x}{r^2 - 2 r \cos x + 1} dx
\end{align*}

so you can't know the value of the integral you are after (##\int_{-\pi}^\pi \dfrac{1}{r^2 - 2 r \cos x + 1} dx##) from the value of ##c_0## of ##u(x)##.

However, you can get the value of the integral you are after from the value of the Fourier coefficient, ##c_0##, of ##v(x) = 1/2 + u(x) e^{ix}##.

Or, as you say, you can use Parseval's on ##u(x)##.