# How do integrate int e^(cos)(t) (sin 2t) dt between a = 0, b = pi?

Apr 10, 2015

Start with ${\int}_{0}^{\pi} {e}^{\cos} \left(t\right) \sin \left(2 t\right) \mathrm{dt} = 2 {\int}_{0}^{\pi} {e}^{\cos} \left(t\right) \sin \left(t\right) \cdot \cos \left(t\right) \mathrm{dt}$

Now by part

$- 2 \int - \sin \left(t\right) {e}^{\cos} \left(t\right) \cdot \cos \left(t\right) \mathrm{dt}$

$u ' = - \sin \left(t\right) {e}^{\cos} \left(t\right)$
$u = {e}^{\cos} \left(t\right)$

$v = \cos \left(t\right)$
$v ' = - \sin \left(t\right)$

You have : $- 2 \left({\left[\cos \left(t\right) \cdot {e}^{\cos} \left(t\right)\right]}_{0}^{\pi} - {\int}_{0}^{\pi} - \sin \left(t\right) {e}^{\cos} \left(t\right) \mathrm{dt}\right)$

$= - 2 {\left[\cos \left(t\right) \cdot {e}^{\cos} \left(t\right) - {e}^{\cos} \left(t\right)\right]}_{0}^{\pi}$

$= \frac{4}{e}$