How to Evaluate $\int e^{2x}\sin(3x)\ dx$?

The key here is to set $v' = e^{2x}$ and integrate by parts twice. This will make $\sin$ become $\cos$ and then $\sin$ again. The same integral will show up on the right side but with a different factor (so it won't cancel out).

Call the integral $I$ and integrate by parts twice to get: \begin{align} I &= \int e^{2x} \sin(3x) \,dx = \frac{1}{2}e^{2x}\sin(3x) - \frac{3}{2} \int e^{2x} \cos(3x) \,dx \\ &= \frac{1}{2}e^{2x}\sin(3x) - \frac{3}{2} \left(\frac{1}{2} e^{2x}\cos(3x) + \frac{3}{2} \int e^{2x} \sin(3x) \,dx\right) \end{align}

Thus: $$ I = \frac{1}{2}e^{2x}\sin(3x) - \frac{3}{2} \left(\frac{1}{2} e^{2x}\cos(3x) + \frac{3}{2} I \right) $$

Solve for $I$ to get:

$$ I = \frac{1}{13}e^{2x}\left(2\sin(3x) -3\cos(3x)\right) $$

HINT

Integrate by parts twice and rearrange.