Answer
$-\frac{1}{13}(3e^{2x}cos3x-2e^{2x}sin3x)+C$
Work Step by Step
Use integration by parts twice.
Set $A=\int e^{2x}sin3xdx$
$u=e^{2x},dv=\int sin3xdx$
$u'=2e^{2x}dx,v=-\frac{1}{3}cos3x$
$A=-\frac{1}{3}e^{2x}cos3x+\int (2e^{2x})(\frac{1}{3}cos3x)dx$
$u=2e^{2x},dv=\int \frac{1}{3}cos3xdx$
$u'=4e^{2x}dx,v=\frac{1}{9}sin3x$
$A=-\frac{1}{3}e^{2x}cos3x+\frac{2}{9}e^{2x}sin3x-\frac{4}{9}\int e^{2x}sin3xdx$
Cross out the $\frac{4}{9}\int e^{2x}sin3xdx$ and replace with $\frac{4}{9}A$ thus:
$A=-\frac{1}{3}e^{2x}cos3x+\frac{2}{9}e^{2x}sin3x-\frac{4}{9}A$
Add $\frac{4}{9}A$ to other side.
$\frac{13}{9}A=-\frac{1}{3}e^{2x}cos3x+\frac{2}{9}e^{2x}sin3x$
Multiply whole equation by $\frac{9}{13}$ to get:
$A=-\frac{3}{13}e^{2x}cos3x+\frac{2}{13}e^{2x}sin3x$
Separate out $-\frac{1}{13}$
$A=-\frac{1}{13}(3e^{2x}cos3x-2e^{2x}sin3x)+C$