How to Evaluate $\int e^{2x}\sin(3x)\ dx$? Integration by parts - unless I'm not thinking straight - doesn't seem to help here.. If I pick either of the functions $e^{2x}$ or $\sin(3x)$ to be $u$ or $dv$, they don't change to anything easier... $e^{2x}$ stays in the form $e^x$ and $\sin(3x)$ flip-flops between $\sin(3x)$ and $\cos(3x)$, neglecting the constant that gets introduced.
So how can I approach solving this? $u$-substitution doesn't seem to be something I can use either.
 A: 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:
$$
I = \frac{1}{13}e^{2x}\left(2\sin(3x) -3\cos(3x)\right)
$$
A: HINT
Integrate by parts twice and rearrange.
A: I just generalize your question a little and put $a$ instead of $2$ and $b$ instead of $3$ in the integrand. You should do the integration by parts twice
$$\eqalign{
  & I = \int {{e^{a\theta }}} \;\sin b\theta \;d\theta   \cr 
  & \,\,\,\, = {1 \over a}{e^{a\theta }}\sin b\theta  - {b \over a}\int {{e^{a\theta }}\cos b\theta \,d\theta }   \cr 
  & \,\,\,\, = {1 \over a}{e^{a\theta }}\sin b\theta  - {b \over a}\left[ {{1 \over a}{e^{a\theta }}\cos b\theta  + {b \over a}\int {{e^{a\theta }}\sin b\theta d\theta} } \right]  \cr 
  & \,\,\,\, = {1 \over a}{e^{a\theta }}\sin b\theta  - {b \over a}\left[ {{1 \over a}{e^{a\theta }}\cos b\theta  + {b \over a}I} \right]  \cr 
  & \,\,\,\, = {1 \over a}{e^{a\theta }}\sin b\theta  - {b \over {{a^2}}}{e^{a\theta }}\cos b\theta  - {{{b^2}} \over {{a^2}}}I \cr} $$
and then solve for $I$. This will result in
$$\eqalign{
  & {{{a^2} + {b^2}} \over {{a^2}}}I = {1 \over a}{e^{a\theta }}\sin b\theta  - {b \over {{a^2}}}{e^{a\theta }}\cos b\theta   \cr 
  & I = {a \over {{a^2} + {b^2}}}{e^{a\theta }}\sin b\theta  - {b \over {{a^2} + {b^2}}}{e^{a\theta }}\cos b\theta  \cr} $$
Or equivalently
$$\boxed{I = {1 \over {{a^2} + {b^2}}}{e^{a\theta }}\left[ {a\sin b\theta  - b\cos b\theta } \right]}$$
