For $z\in \mathbb{C}$ and $n\in \mathbb{N}$, let $f_n(z)$ be the coefficient of $w^n$ in the Laurent series expansion of $e^{\frac{z(w-1/w)}{2}}$. Show that $f_n(z)=\int_{0}^{\pi} cos(n\theta-z sin\theta)d\theta$.
Since the function $g(z)=e^{\frac{z(w-1/w)}{2}}$ is analytic except the origin, I tried to consider the Laurent expansion of $g$ at origin around the curve $r(t)=e^{i\theta}$ where $t\in [0,2\pi]$. Then the coefficient of $w^n$ will be
\begin{align} \frac{1}{2\pi i}\int_{0}^{2\pi} \frac{e^{z/2 [e^{i\theta}-e^{-i\theta}]}}{e^{i(n+1)\theta}}ie^{i\theta}d\theta = \frac{1}{2\pi} \int_{0}^{2\pi} \frac{e^{iz sin\theta}}{e^{in\theta}}d\theta = \frac{1}{2\pi} \int_{0}^{2\pi} e^{i[z sin\theta-n\theta]}d\theta . \end{align}
However, I have no idea how to write the last one as the form $\int_{0}^{\pi} cos(n\theta-z sin\theta)d\theta.$