Without $\theta$

This is nice question and a excellent example of differentiating then integration technique.

Consider , $I(a) = \int_{0}^{\pi} e^{a\cos(\theta)}\cos(a\sin(\theta)) d\theta$ Differnating with respect to $a$ yeild,

$I'(a) = \int_{0}^{\pi} e^{a\cos(\theta)}(\cos(\theta)\cos(a\sin(\theta)) - \sin(\theta)\sin(a\sin(\theta))$

$I'(a) = \frac{1}{a} \Big[e^{a \cos(\theta)}\sin(a\sin(\theta))\Big]_0^{\pi} = 0$

So this is a constant function! . So, $I(1)=I(0) = \pi \Box$

$\begin{aligned} \int_0^\pi e^{\cos{t}}\cos{\left(\sin{t}\right)}\ dt &=&\text{Re}\left(\int_0^\pi e^{\cos{t}}\left(\cos{\left(\sin{t}\right)}+i\sin{\left(\sin{t}\right)}\right)\ dt\right)\\ &=&\text{Re}\left(\int_0^\pi e^{\cos{t}}e^{i\sin{t}}\ dt\right)\\ &=&\text{Re}\left(\int_0^\pi e^{\cos{t}+i\sin{t}}\ dt\right)\\ &=&\text{Re}\left(\int_0^\pi e^{e^{it}}\ dt\right)\\ &=&\text{Re}\left(\int_0^\pi\sum_{k=0}^{\infty}\frac{1}{k!}\left(e^{it}\right)^k\ dt\right)\\ &=&\text{Re}\left(\int_0^\pi 1+\sum_{k=1}^{\infty}\frac{1}{k!}e^{ikt}\ dt\right)\\ &=&\text{Re}\left(\pi-0+\sum_{k=1}^{\infty}\left(\frac{1}{k!}\int_0^\pi e^{ikt}\ dt\right)\right)\\ &=&\text{Re}\left(\pi+\sum_{k=1}^{\infty}\left(\frac{1}{k!}\frac{1}{ik}e^{ikt}\ dt\right)\right)\\ &=&\text{Re}\left(\pi+\sum_{k=1}^\infty\left(\frac{1}{k!}\left(\frac{e^{ik\pi}}{ik}-\frac{e^{ik\cdot0}}{ik}\right)\right)\right)\\ &=&\text{Re}\left(\pi+\sum_{k=1}^\infty\frac{(-1)^k-1}{ik\cdot k!}\right)\\ &=&\text{Re}\left(\pi+\frac{1}{i}\sum_{k=1}^\infty\frac{(-1)^k-1}{k\cdot k!}\right)\\ &=&\text{Re}\left(\pi-i\sum_{k=1}^\infty\frac{(-1)^k-1}{k\cdot k!}\right)\\ &=&\boxed{\pi} \end{aligned}$

Solution Courtesy: @Stefan Stankovic