Integral $\int_0^{2π} e^{e^{ix}} dx$ 
Work out  the integral $$\int_0^{2π}  e^{\large e^{ix}} \, dx.$$ 

I am now stuck with this for $2$ days, so please help! 
Here is my try:
$$I=\int_0^{2π}  e^{\large e^{ix}} dx=\int_0^{2\pi}  e^{\large{\cos x+i\sin x}}dx$$
$$=\int_0^{2\pi}e^{cos x}\left(\cos(\sin x) +i\sin(\sin x)\right) dx$$
$$\overset{\large2\pi -x \rightarrow x}=\int_0^{2\pi} e^{\cos x} \left(\cos (\sin  x)+i\sin(\sin(2\pi- x)\right)dx$$
$$\Rightarrow 2I=\int_0^{2\pi} e^{{\cos x}} \cdot 2\cos (\sin x)dx$$
$$\text{as}\ \sin(2\pi -x )=-\sin x$$
$$\Rightarrow I=\int_0^{2\pi} e^{\cos x}\cos (\sin x)dx$$
$$=\int_0^{2\pi} \left(e^{\cos x}\cos (\sin x)+e^{\cos (\pi-x)}\cos (\sin x)\right)dx$$
$$=\int_0^{\pi} 2 \left(\frac{e^{\cos x}+e^{-\cos x}}{2}\right)\cos(\sin x)dx$$
$$=2\int_0^\pi \operatorname{cosh}(\cos x)\cos(\sin x)dx $$
 A: One trick I learnt a while ago on AoPS was to use Feynman's trick in this case.
Let $f(x)=f =e^{ix}$ for simplicity. And consider: $$I(t)=\int_0^{2π}  e^{\large tf} dx\Rightarrow I'(t)=\int_0^{2\pi} f e^{\large tf}dx$$
But since $(tf)' = \left(te^{ix}\right)'=it e^{ix} =it f$
$$\Rightarrow I'(t)=\frac{1}{it} \int_0^{2\pi} \left(e^{tf}\right)'dx =\frac{e^{tf}}{it}\bigg|_0^{2\pi} $$
Now $e^{ix}$ is periodic with $T=2\pi $  and $e^{2\pi i} = e^0 =1$ so we have:$$I'(t)=\frac{e^t -e^t}{it}=0$$
And since if the derivative of a function is $0$ then the original function must be a constant, this implies that $I(t)$ is simply a constant, and we can set any value we want to obtain the answer.
$$I(t)=I(0)=\int_0^{2\pi} dx=2\pi$$
A: For any $n\in\mathbb{Z}\setminus\{0\}$ we have $\int_{0}^{2\pi}e^{nix}\,dx = 0.$ It follows that
$$ \int_{0}^{2\pi}e^{e^{ix}}\,dx = \sum_{n\geq 0}\frac{1}{n!}\int_{0}^{2\pi}e^{nix}\,dx =\frac{1}{0!}\int_{0}^{2\pi}e^{0ix}\,dx=2\pi.$$
A: Use $z= e^{it}$ and integrate in the circumference of radius 1. Sou your integral becomes 
$$ \int_{C} \frac{e^z}{iz}dz$$ which has a singularity in $z=0$ which is $\frac{1}{i}$ and using the resiude theorem then the integral is $2\pi i \; Res_0(f) = 2\pi i \frac{1}{i} = 2\pi$
A: Make the substitution $z = e^{it}$ to transform the integral into
$$\int_\gamma \frac{e^z}{iz}dz =
\frac{1}{2\pi i}\int_\gamma \frac{2\pi e^z}{z-0}dz$$
where $\gamma$ is the positively oriented closed circular path of radius $1$ around $0$.
Now by Cauchy's Integral Formula this is $2\pi e^0 = 2\pi.$
