# How do I find the integral inte^(2x)*sin(3x)dx ?

Sep 25, 2014

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

Explanation :

$I = \int {e}^{2 x} \cdot \sin \left(3 x\right) \mathrm{dx}$ .........$\left(i\right)$

Using Integration by Parts,

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

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

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

$I = - {e}^{2 x} \cdot \cos \frac{3 x}{3} + \frac{2}{3} {I}_{1}$ .........$\left(i i\right)$

where, ${I}_{1} = \int {e}^{2 x} \cos \left(3 x\right) \mathrm{dx}$

Again, using Integration by Parts,

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

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

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

${I}_{1} = {e}^{2 x} \cdot \sin \frac{3 x}{3} - \frac{2}{3} I$, [using $\left(i\right)$]

now, putting ${I}_{1}$ in $\left(i i\right)$ yields,

$I = - {e}^{2 x} \cdot \cos \frac{3 x}{3} + \frac{2}{3} \left({e}^{2 x} \cdot \sin \frac{3 x}{3} - \frac{2}{3} I\right)$

$I = - {e}^{2 x} \cdot \cos \frac{3 x}{3} + \frac{2}{9} {e}^{2 x} \cdot \sin \left(3 x\right) - \frac{4}{9} I$

$I + \frac{4}{9} I = - {e}^{2 x} \cdot \cos \frac{3 x}{3} + \frac{2}{9} {e}^{2 x} \cdot \sin \left(3 x\right)$

$\frac{13}{9} I = {e}^{2 x} / 3 \left(\frac{2}{3} \sin \left(3 x\right) - \cos \left(3 x\right)\right)$

$I = \frac{3}{13} {e}^{2 x} \left(\frac{2}{3} \sin \left(3 x\right) - \cos \left(3 x\right)\right)$