The integral is:
$$\int_{0}^{2\pi}e^{-\sin\theta}\cos(\theta+\cos\theta)d\theta$$
My first thoughts were to use the Cauchy-Riemann equations in polar coordinates to show that the function is regular over the integration domain and therefore applying Cauchy's Theorem directly. However, since the function is real, it's imaginary part, $v$, equals $0$. With that, the CR equations don't seem to hold.
Any ideas/hints?
EDIT: I forgot to add a minus sign.
On
Let $f(\theta)=e^{-\sin\theta}\cos(\theta+\cos\theta)$. Note that $$ f(\pi-\theta)=e^{-\sin\theta}\cos(\pi-\theta-\cos\theta)=-e^{-\sin\theta}\cos(\theta+\cos\theta)=-f(\theta). $$ So $$ \int_{0}^{2\pi}f(\theta)\,d\theta=\int_{-\pi}^{\pi}f(\theta)\,d\theta+\int_{\pi}^{3\pi}f(\theta)\,d\theta=\int_{-\pi}^{\pi}[f(\theta)+f(\pi-\theta)]\,d\theta=0. $$
$\newcommand{\d}{\,\mathrm{d}}$You may rewrite: $$\begin{align}\int_0^{2\pi}e^{-\sin\theta}\cos(\theta+\cos(\theta))\d\theta&=\frac{1}{2}\int_0^{2\pi}e^{i\theta}e^{i\cos\theta-\sin\theta}\d \theta+\frac{1}{2}\int_0^{2\pi}e^{-i\theta}e^{-(i\cos\theta+\sin\theta)}\d \theta\\&=\frac{1}{2i}\oint_{|z|=1}e^{iz}\d z+\frac{1}{2i}\oint_{|z|=1}e^{-iz}\d z\\&=-i\oint_{|z|=1}\cos(z)\d z\\&=0\quad\text{by holomorphy, Cauchy Integral Theorem}\end{align}$$
Where the contours $z=e^{it}$ and $z=e^{-it}$ were substituted (a double negative preserves orientation in the second integral).