# How do you find the antiderivative of (e^(2x))sin3x?

Aug 14, 2017

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

#### Explanation:

We could use a traditional double application of Integration By Parts. Here is a slightly different approach.

Let:

$s = {e}^{2 x} \sin 3 x \setminus \setminus \setminus \setminus$ and ${I}_{s} = \int {e}^{2 x} \sin 3 x$
$c = {e}^{2 x} \cos 3 x \setminus \setminus \setminus \setminus$ and ${I}_{c} = \int {e}^{2 x} \cos 3 x$

Differentiating wrt $x$ we get:

$\frac{\mathrm{ds}}{\mathrm{dx}} = {e}^{2 x} \left(\frac{d}{\mathrm{dx}} \sin 3 x\right) + \left(\frac{d}{\mathrm{dx}} {e}^{2 x}\right) \sin 3 x$
$\setminus \setminus \setminus \setminus \setminus \setminus = 3 {e}^{2 x} \cos 3 x + 2 {e}^{2 x} \sin 3 x$

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

Now integrate the above results:

$\int \setminus \frac{\mathrm{ds}}{\mathrm{dx}} \setminus \mathrm{dx} = \int \setminus 3 {e}^{2 x} \cos 3 x + 2 {e}^{2 x} \sin 3 x \setminus \mathrm{dx}$
$\implies s = 3 {I}_{c} + 2 {I}_{s}$ ... [A]

$\int \setminus \frac{\mathrm{dc}}{\mathrm{dx}} \setminus \mathrm{dx} = \int \setminus - 3 {e}^{2 x} \sin 3 x + 2 {e}^{2 x} \cos 3 x \setminus \mathrm{dx}$
$\implies c = - 3 {I}_{s} + 2 {I}_{c}$ ... [B]

3Eq [A] + 2Eq [B}:

$3 s + 2 c = 9 {I}_{c} + 6 {I}_{s} - 6 {I}_{s} + 4 {I}_{c}$
$\therefore 3 s + 2 c = 13 {I}_{c}$
$\therefore {I}_{c} = \frac{1}{13} \left(3 s + 2 c\right)$

From [A] we also get:

$s = \frac{3}{13} \left(3 s + 2 c\right) + 2 {I}_{s}$
$\therefore s = \frac{9}{13} s + \frac{6}{13} c + 2 {I}_{s}$
$\therefore {I}_{s} = \frac{1}{13} \left(2 s - 3 c\right)$

Hence we get the two results:

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

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