With $\theta$

Use substitution first:

Let $x=\cos \theta$

$\int{e^{\cos \theta} \cos \theta\, \sin \theta \, d \theta} = - \int{e^{x} x\, \, d x}$

Now, integrate by parts:

$- \int{x\,e^x\,dx} = - (x\,e^x - e^x) = e^x - x\,e^x$

Ignore the constant of integration.

$\int^{\pi}_{0}{e^{\cos \theta} \cos \theta\, \sin \theta \, d \theta} = e^x - x\,e^x = e^{\cos \theta} - \cos \theta \, e^{\cos \theta}$

To put back the limits, we insert $\theta = \pi$ and simplify to get:

$\int^{\pi}_{0}{e^{\cos \theta} \,\cos \theta\, \sin \theta \, d \theta} = \frac{2}{e}$

Now, do the rest yourself..... I must go for lunch

$I=\int _{ 0 }^{ \pi }{ { e }^{ cos\theta }cos\theta \quad sin\theta } d\theta \\ put\quad cos\theta =t\\ then\quad -sin\theta d\theta =dt\\ when\quad \theta =\pi \quad t=-1\\ \quad \quad \quad \quad \quad \theta =0\quad t=1\\ then\quad I=-\int _{ 1 }^{ -1 }{ { e }^{ t } } t\quad dt\\ \quad \quad \quad \quad \quad =\int _{ -1 }^{ 1 }{ { e }^{ t } } t\quad dt\\ \quad \quad \quad \quad \quad =\frac { 2 }{ e }$

$\displaystyle \int_{0}^{\pi} e^{cos \theta}\cos\theta\sin\theta d\theta$ The sustitution that makes this easier is $cos\theta = x => d\theta = \frac{dx}{-sin\theta}$ . Therefore $\displaystyle \int_{0}^{\pi} e^{cos\theta}\cos\theta\sin\theta d\theta = \displaystyle \int_{-1}^{1} e^{x}x dx$ $(e^{x}x]_{-1}^{1}-\displaystyle \int_{-1}^{1}e^{x} dx$ $=e+\frac{1}{e} -e+\frac{1}{e}$