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:
$$I=\int_0^{2π}  e^{\large e^{ix}} dx=\int_0^{2\pi}  e^{\large{\cos x+i\sin x}}dx$$
$$=\int_0^{2\pi}e^{cos x}\left(\cos(\sin x) +i\sin(\sin x)\right) dx$$
$$\overset{\large2\pi -x \rightarrow x}=\int_0^{2\pi} e^{\cos x} \left(\cos (\sin  x)+i\sin(\sin(2\pi- x)\right)dx$$
$$\Rightarrow 2I=\int_0^{2\pi} e^{{\cos x}} \cdot 2\cos (\sin x)dx$$
$$\text{as}\ \sin(2\pi -x )=-\sin x$$
$$\Rightarrow I=\int_0^{2\pi} e^{\cos x}\cos (\sin x)dx$$
$$=\int_0^{2\pi} \left(e^{\cos x}\cos (\sin x)+e^{\cos (\pi-x)}\cos (\sin x)\right)dx$$
$$=\int_0^{\pi} 2 \left(\frac{e^{\cos x}+e^{-\cos x}}{2}\right)\cos(\sin x)dx$$
$$=2\int_0^\pi \operatorname{cosh}(\cos x)\cos(\sin x)dx $$

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 $2\pi i \; Res_0(f) = 2\pi i \frac{1}{i} = 2\pi$
