Fourier series of $\arccos(\lambda\cos x$)

Question

Please help with finding the (cosine) Fourier series of

$$f(x)=\arccos(\lambda\cos x),$$ where $\lambda$ is a real number, $|\lambda|<1$.

It is easy to find that $c_0=\pi/2$ and $c_{2n}=0$, but I do not see a way to integrate the expression for odd $n$. Any hint is welcome.

Answer 1

One idea: We know the arcus cosine series $$ \frac{d}{du}\arccos = -\frac1{\sqrt{1-u^2}}=-\sum\binom{2n}{n}\frac{u^{2n}}{4^n} \\ \implies \arccos u=\frac\pi2-\sum\binom{2n}{n}\frac{u^{2n+1}}{4^n(2n+1)} $$ Now insert $u=λ\cos x=\fracλ2(e^{ix}+e^{-ix})$, which is permissible as $|u|<1$ is inside the series radius of convergence, apply the binomial theorem and sort the powers of $e^{ix}$.