# How do you integrate int e^x sin x dx  using integration by parts?

Jan 7, 2016

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

#### Explanation:

Say $\mathrm{dv} = {e}^{x}$ so $v = {e}^{x}$, $u = \sin \left(x\right)$ so $\mathrm{du} = \cos \left(x\right)$

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

Say $\mathrm{dv} = {e}^{x}$ so $v = {e}^{x}$, $u = \cos \left(x\right)$ so $\mathrm{du} = - \sin \left(x\right)$

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