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

1 Answer
Aug 27, 2016

Answer:

#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))#