Power series expansion of $e^{\frac{c}{2}(z-1/z)}$

Question

Show that

\begin{equation} e^{\frac{c}{2}(z-1/z)}=\sum_{n=-\infty}^{\infty}a_nz^n \end{equation}

where

\begin{equation} a_n:=\frac{1}{2\pi}\int_0^{2\pi}\cos(n\theta-c\sin(\theta))d\theta \end{equation}