complex analysis - Laurent Series Coefficients Problem.

I am struggling with a question regarding Laurent's Theorem and the coefficients of a Laurent series. The question is attached above. I know the general formula for coefficients. I have set $z=e^{i \theta}$. I have plugged in also $f(z)$ into the coefficient formula given by Laurent's Theorem. I however, cannot get the answer to be or look like the integral for the coefficients with the cosine etc, given in the question, which is what we are trying to prove or show. Do you have any pointers or suggestions? That would really help. Thank you.

Answer

The function $e^{z+\frac{1}{z}}$ is holomorphic everywhere in $0 < |z| < \infty$ and, therefore, has a Laurent series expansions
$$
e^{z+\frac{1}{z}} = \sum_{n=-\infty}^{\infty}a_n z^n, \;\;\; 0 < |z| < \infty.

$$
The Laurent series coefficients $a_n$ are given by
\begin{align}
a_n&=\frac{1}{2\pi i}\oint_{|z|=1}e^{z+\frac{1}{z}}\frac{1}{z^{n+1}}dz \\
&= \frac{1}{2\pi }\int_{-\pi}^{\pi}e^{e^{i\theta}+e^{-i\theta}}e^{-i(n+1)\theta}e^{i\theta}d\theta \\
&= \frac{1}{2\pi }\int_{-\pi}^{\pi}e^{2\cos\theta}e^{-in\theta}d\theta \\
&= \frac{1}{2\pi }\left(\int_{-\pi}^{0}+\int_{0}^{\pi}\right)e^{2\cos\theta}e^{-in\theta}d\theta \\
&= \frac{1}{2\pi }\left(
- \int_{\pi}^{0}e^{2\cos(-\theta)}e^{in\theta}d\theta+\int_{0}^{\pi}e^{2\cos\theta}e^{-in\theta}d\theta\right) \\
&= \frac{1}{2\pi }\int_{0}^{\pi}e^{2\cos\theta}(e^{in\theta}+e^{-in\theta})d\theta \\

&= \frac{1}{\pi}\int_{0}^{\pi}e^{2\cos\theta}\cos(n\theta)d\theta.
\end{align}