$\displaystyle \int_0^\pi e^{cos(t)}sin(2t)dt$

=

$\displaystyle 2\int_0^\pi e^{cos(t)}sin(t)cos(t)dt$

$\displaystyle u = cos(t) $

$\displaystyle du = -sin(t)$

When $\displaystyle t = \pi $ $\displaystyle u = -1 $

When $\displaystyle t = 0$ $\displaystyle u = 1 $

$\displaystyle -2\int_1^{-1} ue^u du$

Then integration by parts, differentiating $\displaystyle u $and integrating $\displaystyle e^u $ i end up with

$\displaystyle -2e^u(u-1)$ from 1 to -1

Doing that I end up with 0 after plugging $\displaystyle cos(t) $ back in for the u. but my calculator tells me the answer is $\displaystyle 4e^-1$

What am I doin wrong, I know I'm close...i think :P

PS: best part is, if i just substitute the -1 and 1 directly into the above equation...i get the right answer

EDIT: ....im suppose to put -1 and 1 in aren't i....shrug haha