# How do you find the antiderivative of e^(2x)*sin(e^x)dx?

Jan 17, 2017

$\int {e}^{2 x} \sin \left({e}^{x}\right) \mathrm{dx} = - {e}^{x} \cos \left({e}^{x}\right) + \sin \left({e}^{x}\right) + C$

#### Explanation:

$I = \int {e}^{2 x} \sin \left({e}^{x}\right) \mathrm{dx}$

Let $t = {e}^{x}$. This implies that $\mathrm{dt} = {e}^{x} \mathrm{dx}$. Before substituting, note that ${e}^{2 x} = {e}^{x} \left({e}^{x}\right)$.

$I = \int {e}^{x} \sin \left({e}^{x}\right) \left({e}^{x} \mathrm{dx}\right) = \int t \sin \left(t\right) \mathrm{dt}$

To do this, use integration by parts. This takes the form $\int u \mathrm{dv} = u v - \int v \mathrm{du}$. Let:

$\left\{\begin{matrix}u = t \text{ "=>" "du=dt \\ dv=sin(t)dt" "=>" } v = - \cos \left(t\right)\end{matrix}\right.$

Then:

$I = u v - \int v \mathrm{du} = - t \cos \left(t\right) - \int \left(- \cos \left(t\right)\right) \mathrm{dt}$

$I = - t \cos \left(t\right) + \int \cos \left(t\right) \mathrm{dt} = - t \cos \left(t\right) + \sin \left(t\right)$

Returning to the original variable using $t = {e}^{x}$:

$I = - {e}^{x} \cos \left({e}^{x}\right) + \sin \left({e}^{x}\right) + C$