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

Feb 2, 2017

I got: $\frac{9}{13} \left[{e}^{3 x} / 3 \cos \left(2 x\right) + \frac{2}{9} {e}^{3 x} \sin \left(2 x\right)\right] + c$
BUT check my maths.

#### Explanation:

Ok let us try:
$\int {e}^{3 x} \cos \left(2 x\right) \mathrm{dx} = {e}^{3 x} / 3 \cos \left(2 x\right) - \int \left({e}^{3 x} / 3\right) \left(- 2 \sin \left(2 x\right)\right) \mathrm{dx} = {e}^{3 x} / 3 \cos \left(2 x\right) + \frac{2}{3} \int \left({e}^{3 x}\right) \sin \left(2 x\right) \mathrm{dx} =$
again:
$= {e}^{3 x} / 3 \cos \left(2 x\right) + \frac{2}{3} \left[{e}^{3 x} / 3 \sin \left(2 x\right) - \int \left({e}^{3 x} / 3\right) \left(2 \cos \left(2 x\right)\right) \mathrm{dx}\right]$

So basically we get:

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

Now a trick....
Let us take $- \frac{4}{9} \int {e}^{3 x} \left(2 \cos \left(2 x\right)\right) \mathrm{dx}$ to the left of the $=$ sign:
$\int {e}^{3 x} \cos \left(2 x\right) \mathrm{dx} + \frac{4}{9} \int {e}^{3 x} \left(2 \cos \left(2 x\right)\right) \mathrm{dx} = {e}^{3 x} / 3 \cos \left(2 x\right) + \frac{2}{9} {e}^{3 x} \sin \left(2 x\right)$
add the two integral on the left:
$\frac{13}{9} \int {e}^{3 x} \left(2 \cos \left(2 x\right)\right) \mathrm{dx} = {e}^{3 x} / 3 \cos \left(2 x\right) + \frac{2}{9} {e}^{3 x} \sin \left(2 x\right)$
and:
$\int {e}^{3 x} \left(2 \cos \left(2 x\right)\right) \mathrm{dx} = \frac{9}{13} \left[{e}^{3 x} / 3 \cos \left(2 x\right) + \frac{2}{9} {e}^{3 x} \sin \left(2 x\right)\right] + c$