Consider the function $$g:[0,2\pi]\to \mathbb{R}, \quad x\mapsto2e^{2\cos x}\cos(2 \sin x) -1$$
I would like to find its Fourier coefficients.
Since $g$ is an even function, the Fourier coefficients are given as $$\alpha_k = \frac{1}{\pi} \int_0^{\pi} g(x) \cos(kx) dx \quad (k\in\mathbb{N}).$$
I'm not sure if this is helpful but I've written $g$ as follows: $g(x) = e^{2e^{ix}} + e^{2e^{-ix}}-1$.
Please help me.
Hint. You may observe that, for $x \in \mathbb{R}$, we have $$ \begin{align} e^{2\cos x}\cos(2 \sin x)&=\Re \left(e^{2\cos x}e^{2i\sin x}\right)\\\\ &=\Re \left(e^{2e^{ix}}\right)\\\\ &=\Re \left(\sum_{n=0}^{\infty}\frac{\left(2e^{ix}\right)^n}{n!}\right)\\\\ &=\Re \left(\sum_{n=0}^{\infty}\frac{2^ne^{inx}}{n!}\right)\\\\ &=\sum_{n=0}^{\infty}\frac{2^n\cos (nx)}{n!}\\\\ \end{align} $$ then by the uniqueness of the Fourier coefficients you obtain $$ b_n = \frac{2^{n+1}}{n!}\pi, \quad n=1,2,3,\ldots . $$ with $$ b_n = \int_0^{2\pi} e^{2\cos x}\cos(2 \sin x) \cos(nx) dx. $$