Calculating complex integral for different values of $n$.

Solution 1:

$$\newcommand{\Re}{\operatorname{Re}}\newcommand{\Res}{\operatorname*{Res}} \begin{align} \int_0^{2\pi}e^{{}-\cos(\theta)}\cos(n\theta+\sin(\theta))\,\mathrm{d}\theta &=\Re\left(\int_0^{2\pi}e^{{}-\cos(\theta)-in\theta-i\sin(\theta)}\,\mathrm{d}\theta\right)\tag1\\ &=\Re\left(\int_0^{2\pi}e^{-e^{i\theta}}\,e^{-in\theta}\,\mathrm{d}\theta\right)\tag2\\ &=\Re\left(\frac1i\int_0^{2\pi}e^{-e^{i\theta}}\,e^{-i(n+1)\theta}\,\mathrm{d}e^{i\theta}\right)\tag3\\ &=\Re\left(\frac1i\oint_{|z|=1}e^{-z}z^{-n-1}\,\mathrm{d}z\right)\tag4\\[3pt] &=2\pi\Re\left(\Res_{z=0}\left(e^{-z}z^{-n-1}\right)\right)\tag5\\[3pt] &=2\pi\frac{(-1)^n}{n!}[n\ge0]\tag6 \end{align} $$ Explanation:
$(1)$: $\cos(z)=\Re\left(e^{-iz}\right)$
$(2)$: $\cos(\theta)+i\sin(\theta)=e^{i\theta}$
$(3)$: $\mathrm{d}e^{i\theta}=ie^{i\theta}\,\mathrm{d}\theta$
$(4)$: substitute $z=e^{i\theta}$
$(5)$: Residue Theorem
$(6)$: use the power series for $e^z$ to get $\left[z^n\right]e^{-z}$