integrate $\int_0^{2\pi} e^{\cos \theta} \cos( \sin \theta) d\theta$ How to integrate
$ 1)\displaystyle \int_0^{2\pi} e^{\cos \theta} \cos( \sin \theta) d\theta$
$ 2)\displaystyle \int_0^{2\pi} e^{\cos \theta} \sin ( \sin \theta) d\theta$
 A: $$\int_0^{2\pi} e^{\cos\theta}\cos(\sin\theta)d\theta=\Re\left(\int_0^{2\pi} e^{e^{i\theta}}d\theta\right)$$
$$I(\lambda)=\int_0^{2\pi} e^{\lambda e^{i\theta}}d\theta$$
$$I'(\lambda)=\frac{1}{i\lambda}\int_0^{2\pi} i\lambda e^{i\theta}e^{\lambda e^{i\theta}}d\theta =\frac{1}{i\lambda}\bigg[e^{\lambda e^{i\theta}}\bigg]_0^{2\pi}=0$$
Hence:
$$I(\lambda)=\mathcal{C}$$
Taking $\lambda =0$ we have $\mathcal{C}=2\pi$ so:
$$\int_0^{2\pi} e^{\cos\theta}\cos(\sin\theta)d\theta=2\pi$$
Similarly, since $\Im\, (2\pi)=0:$
$$\int_0^{2\pi} e^{\cos\theta}\sin(\sin\theta)d\theta=0$$
A: Hint: If the first integral is called $I$ and the second is called $J$
Consider $I+iJ$
A: Let $\gamma$ be the unitary circumference positively parametrized going around just once.
Consider $\displaystyle \int _\gamma \frac{e^z}{z}\,dz$.
On the one hand $$\begin{align}
\int _\gamma \frac{e^z}{z}\mathrm dz&=\int \limits_0^{2\pi}\frac{e^{e^{i\theta}}}{e^{i\theta}}ie^{i\theta}\mathrm d\theta\\
&=i\int _0^{2\pi}e^{\cos (\theta)+i\sin (\theta )}\mathrm d\theta\\
&=i\int _0^{2\pi}e^{\cos (\theta )}[\cos (\sin (\theta))+i\sin (\sin (\theta))\textbf{]}\mathrm d\theta.
\end{align}$$
On the other hand Cauchy's integral formula gives you: $\displaystyle \int _\gamma \frac{e^z}{z}\mathrm dz=2\pi i$.
$\large \color{red}{\text{FINISH HIM!}}$
A: Hint:
$$\cos(\sin{\theta})=\frac{e^{i\sin{\theta}}+e^{-i\sin{\theta}}}{2}$$
so
$$e^{\cos{\theta}}\cos(\sin{\theta})=\frac{1}{2}\left(e^{\cos{\theta}+i\sin{\theta}}+e^{\cos{\theta}-i\sin{\theta}}\right)=\frac{1}{2}\left(e^z+e^{\bar{z}} \right), \\
d{\theta}=\frac{dz}{iz},$$
thus
$$
\int\limits_0^{2\pi} e^{\cos{\theta}}\cos(\sin{\theta})d{\theta}=\frac{1}{2}\left(\oint\limits_{\gamma}{\frac{e^{z}dz}{iz}}+\oint\limits_{\gamma}{\frac{e^{\bar{z}}dz}{iz}} \right),
$$
where $\gamma$ denotes the unit circle.
Added:
The last integral can be transformed as follows
$$\oint\limits_{\gamma}{\frac{e^{\bar{z}}dz}{iz}}=\int\limits_{0}^{2\pi}{{e^{\rho{e^{-i\theta}}}d{\theta}}}=\left|\matrix{\varphi=2\pi-\theta  \\
d{\theta}=-d{\varphi}
}\right|= \\
=-\int\limits_{2\pi}^{0}{{e^{\rho{e^{i(\varphi-2\pi)}}}d{\varphi}}}=\int\limits_{0}^{2\pi}{{e^{\rho{e^{i\varphi}}}d{\varphi}}}=\oint\limits_{\gamma}{\frac{e^{w}dw}{iw}}.$$
