Question

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

Answer 1

`e^(2x)/13(2sin(3x)-3cos(3x))+C`

Explanation:

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

`int e^(2x)cos(3x)dx + i int e^(2x)*sin(3x) dx =int e^(2x) e^(i3x)dx`

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

but

`int e^(2x) e^(i3x)dx = int e^(2x+i3x)dx = int e^((2+3i)x)dx`

which gives

`int e^((2+3i)x)dx = e^((2+3i)x)/(2+3i)=(e^((2+3i)x)(2-3i))/13 = (2-3i)/13 e^(2x) e^(i3x)=(2-3i)/13e^(2x)(cos(3x)+i sin(3x))` with imaginary part given by

`e^(2x)/13(2sin(3x)-3cos(3x))`