integrate $\int_0^{2\pi} e^{\cos \theta} \cos( \sin \theta) d\theta$ [closed]

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$

Solution 1:

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: $\displaystyle \int _\gamma \frac{e^z}{z}\mathrm dz=2\pi i$.

$\large \color{red}{\text{FINISH HIM!}}$

Solution 2:

$$\int_0^{2\pi} e^{\cos\theta}\cos(\sin\theta)d\theta=\Re\left(\int_0^{2\pi} e^{e^{i\theta}}d\theta\right)$$

$$I(\lambda)=\int_0^{2\pi} e^{\lambda e^{i\theta}}d\theta$$

$$I'(\lambda)=\frac{1}{i\lambda}\int_0^{2\pi} i\lambda e^{i\theta}e^{\lambda e^{i\theta}}d\theta =\frac{1}{i\lambda}\bigg[e^{\lambda e^{i\theta}}\bigg]_0^{2\pi}=0$$

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 $\Im\, (2\pi)=0:$

$$\int_0^{2\pi} e^{\cos\theta}\sin(\sin\theta)d\theta=0$$

Solution 3:

Hint: If the first integral is called $I$ and the second is called $J$

Consider $I+iJ$