# How do you find the indefinite integral of int e^(2x)sin(7x)dx?

Sep 20, 2015

$\frac{1}{53} {e}^{2 x} \left(2 \sin 7 x - 7 \cos 7 x\right) + C$

#### Explanation:

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

$I = \int {e}^{2 x} \sin 7 x \mathrm{dx}$

${e}^{2 x} = u \implies 2 {e}^{2 x} \mathrm{dx} = \mathrm{du}$
$\mathrm{dv} = \sin 7 x \mathrm{dx} \implies v = \int \sin 7 x \mathrm{dx} = - \frac{1}{7} \cos 7 x$

$I = - \frac{1}{7} {e}^{2 x} \cos 7 x + \frac{2}{7} \int {e}^{2 x} \cos 7 x \mathrm{dx}$

Again:

${e}^{2 x} = u \implies 2 {e}^{2 x} \mathrm{dx} = \mathrm{du}$

$\mathrm{dv} = \cos 7 x \mathrm{dx} \implies v = \int \cos 7 x \mathrm{dx} = \frac{1}{7} \sin 7 x$

$I = - \frac{1}{7} {e}^{2 x} \cos 7 x + \frac{2}{7} \left[\frac{1}{7} {e}^{2 x} \sin 7 x - \frac{2}{7} \int {e}^{2 x} \sin 7 x \mathrm{dx}\right]$

$I = - \frac{1}{7} {e}^{2 x} \cos 7 x + \frac{2}{49} {e}^{2 x} \sin 7 x - \frac{4}{49} \int {e}^{2 x} \sin 7 x \mathrm{dx}$

$I = - \frac{1}{7} {e}^{2 x} \cos 7 x + \frac{2}{49} {e}^{2 x} \sin 7 x - \frac{4}{49} I$

$I + \frac{4}{49} I = - \frac{1}{7} {e}^{2 x} \cos 7 x + \frac{2}{49} {e}^{2 x} \sin 7 x$

$\frac{53}{49} I = - \frac{1}{7} {e}^{2 x} \cos 7 x + \frac{2}{49} {e}^{2 x} \sin 7 x$

$I = - \frac{7}{53} {e}^{2 x} \cos 7 x + \frac{2}{53} {e}^{2 x} \sin 7 x + C$

$I = \frac{1}{53} {e}^{2 x} \left(2 \sin 7 x - 7 \cos 7 x\right) + C$