Answer
$\frac{4}{e}$
Work Step by Step
First, we use a trig double angle property: $sin(2t)=2cos(t)sin(t)$
Our new integral is now $\int_{0}^{\pi}2e^{cos(t)}cos(t)sin(t)dt$.
Substitute $a=cos(t)$.
$da=-sin(t)dt$
Change the limits by plugging them into $t$.
$-\int_{1}{-1}2e^ada=-2\int_{1}^{-1}e^aa\thinspace da$
Integration by parts:
$u=a$, $dv=e^ada$
$du=da$, $v=e^a$
$-2\left[ae^a-\int e^a\thinspace da\right]_{1}^{-1}$
$=-2\left[ae^a-e^a\right]_{1}^{-1}$
$=\frac{4}{e}$
