# How do you find the antiderivative of e^(2x)*sin(3x) dx?

Aug 27, 2016

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

#### Explanation:

This kind of integral can be easily handled by inmersion in the complex domain. Solving instead

$\int {e}^{2 x} \cos \left(3 x\right) \mathrm{dx} + i \int {e}^{2 x} \cdot \sin \left(3 x\right) \mathrm{dx} = \int {e}^{2 x} {e}^{i 3 x} \mathrm{dx}$

(We are using de Moivre's identity e^(i phi) = cos(phi)+i sin(phi))

but

$\int {e}^{2 x} {e}^{i 3 x} \mathrm{dx} = \int {e}^{2 x + i 3 x} \mathrm{dx} = \int {e}^{\left(2 + 3 i\right) x} \mathrm{dx}$

which gives

$\int {e}^{\left(2 + 3 i\right) x} \mathrm{dx} = {e}^{\left(2 + 3 i\right) x} / \left(2 + 3 i\right) = \frac{{e}^{\left(2 + 3 i\right) x} \left(2 - 3 i\right)}{13} = \frac{2 - 3 i}{13} {e}^{2 x} {e}^{i 3 x} = \frac{2 - 3 i}{13} {e}^{2 x} \left(\cos \left(3 x\right) + i \sin \left(3 x\right)\right)$ with imaginary part given by

${e}^{2 x} / 13 \left(2 \sin \left(3 x\right) - 3 \cos \left(3 x\right)\right)$