# How do you find the integral of f(x)=e^(2x)sin3x using integration by parts?

Nov 5, 2015

$\frac{1}{13} {e}^{2 x} \left(2 \sin \left(3 x\right) - 3 \cos \left(3 x\right)\right) + c$

#### Explanation:

So, with integration by parts we state that:

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

Let's use ${e}^{2 x}$ as our u function and $\sin 3 x$ as our v' function, leaving us with:

$u = {e}^{2 x}$
$u ' = 2 {e}^{2 x}$

$v = - \frac{1}{3} \cos 3 x$
$v ' = \sin 3 x$

Now let's plug this back into our original formula:

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

Let's simplify it a bit more:

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

We have to apply integration by parts yet again, using:

$u = {e}^{2 x}$
$u ' = 2 {e}^{2 x}$

$v ' = \cos \left(3 x\right)$
$v = \frac{1}{3} \sin \left(3 x\right)$

Ultimately giving us:

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

Now, if we isolate $\int \sin \left(3 x\right) {e}^{2 x} \mathrm{dx}$ from this equation we'll end up with:

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

I'm taking Calc 2 as well, so if anyone has a correction please don't hesitate! Also I'm running late for class so I kinda glossed over the last step, I can explain further if needed!