# How do you integrate int e^(2x)sin(3x) by integration by parts method?

Dec 28, 2017

$\int {e}^{2 x} \sin \left(3 x\right) \mathrm{dx} = - \frac{3}{13} {e}^{2 x} \cos \left(3 x\right) + \frac{2}{13} {e}^{2 x} \sin \left(3 x\right)$

#### Explanation:

$\int {e}^{2 x} \sin \left(3 x\right) \mathrm{dx}$

if $u = {e}^{2 x}$ and $\mathrm{dv} = \sin \left(3 x\right) \mathrm{dx}$, then $\mathrm{du} = 2 {e}^{2 x} \mathrm{dx}$ and $v = - \frac{1}{3} \cos \left(3 x\right)$.

using integration by parts, the integral becomes:

$- \frac{1}{3} {e}^{2 x} \cos \left(3 x\right) - \int - \frac{1}{3} \cdot 2 {e}^{2 x} \cos \left(3 x\right) \mathrm{dx}$
$= - \frac{1}{3} {e}^{2 x} \cos \left(3 x\right) + \frac{2}{3} \int {e}^{2 x} \cos \left(3 x\right) \mathrm{dx}$

integrate by parts again:

$u = {e}^{2 x}$ and $\mathrm{dv} = \cos \left(3 x\right) \mathrm{dx}$, which means $\mathrm{du} = 2 {e}^{2 x} \mathrm{dx}$ and $v = \frac{1}{3} \sin \left(3 x\right)$

the integral is now:
$= - \frac{1}{3} {e}^{2 x} \cos \left(3 x\right) + \frac{2}{3} \left(\frac{1}{3} {e}^{2 x} \sin \left(3 x\right) - \int 2 \cdot \frac{1}{3} {e}^{2 x} \sin \left(3 x\right) \mathrm{dx}\right)$
$= - \frac{1}{3} {e}^{2 x} \cos \left(3 x\right) + \frac{2}{9} {e}^{2 x} \sin \left(3 x\right) - \frac{2}{3} \int \frac{2}{3} {e}^{2 x} \sin \left(3 x\right) \mathrm{dx}$
$= - \frac{1}{3} {e}^{2 x} \cos \left(3 x\right) + \frac{2}{9} {e}^{2 x} \sin \left(3 x\right) - \frac{4}{9} \int {e}^{2 x} \sin \left(3 x\right) \mathrm{dx}$

remember that this expression is equal to the original integral, meaning:

$\int {e}^{2 x} \sin \left(3 x\right) \mathrm{dx} = - \frac{1}{3} {e}^{2 x} \cos \left(3 x\right) + \frac{2}{9} {e}^{2 x} \sin \left(3 x\right) - \frac{4}{9} \int {e}^{2 x} \sin \left(3 x\right) \mathrm{dx}$

solve for $\int {e}^{2 x} \sin \left(3 x\right) \mathrm{dx}$:

$\frac{13}{9} \int {e}^{2 x} \sin \left(3 x\right) \mathrm{dx} = - \frac{1}{3} {e}^{2 x} \cos \left(3 x\right) + \frac{2}{9} {e}^{2 x} \sin \left(3 x\right)$
$\int {e}^{2 x} \sin \left(3 x\right) \mathrm{dx} = \frac{9}{13} \left(- \frac{1}{3} {e}^{2 x} \cos \left(3 x\right) + \frac{2}{9} {e}^{2 x} \sin \left(3 x\right)\right)$
$\int {e}^{2 x} \sin \left(3 x\right) \mathrm{dx} = - \frac{3}{13} {e}^{2 x} \cos \left(3 x\right) + \frac{2}{13} {e}^{2 x} \sin \left(3 x\right)$