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

Apr 11, 2018

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

#### Explanation:

Integrate by parts:

$\int {e}^{7 x} \sin 2 x \mathrm{dx} = \int \sin 2 x d \left({e}^{7 x} / 7\right)$

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

and then again:

$\int {e}^{7 x} \sin 2 x \mathrm{dx} = \frac{{e}^{7 x} \sin 2 x}{7} - \frac{2}{7} \int \cos 2 x d \left({e}^{7 x} / 7\right)$

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

The same integral now appears on both sides of the equation and we can solve for it:

$\frac{53}{49} \int {e}^{7 x} \sin 2 x \mathrm{dx} = \frac{{e}^{7 x} \sin 2 x}{7} - \frac{2 {e}^{7 x} \cos 2 x}{49} + C$

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