<s>[INST] How can I solve $\int e^{2\theta} \sin(3\theta)\, d\theta$ with integration by parts? $\int e^{2\theta}\sin(3\theta)d\theta$ seems to be leading me in circles. The integral I get when I use integration by parts, $\int e^{2\theta}\cos(3\theta)d\theta$ just leads me back to $\int e^{2\theta}\sin(3\theta)d\theta$. I am not sure how to solve it.
My Steps:
$\int e^{2\theta}\sin(3\theta)d\theta$
Let $u = \sin(3\theta)$ and $dv=e^{2\theta}d\theta$
Then $du = 3\cos(3\theta)d\theta$ and $v = \frac{1}{2}e^{2\theta}$
\begin{align*}
\int e^{2\theta} \sin(3 \theta)d\theta &= \frac{1}{2} e^{2\theta}\sin(3\theta) - \int\frac{1}{2}e^{2\theta}3\cos(3\theta)d\theta\\
&=e^{2\theta}\sin(3\theta) - \frac{3}{2}\int e^{2\theta}\cos(3\theta)d\theta\\
\end{align*}

$\int e^{2\theta}\cos(3\theta)d\theta$
Let $u = \cos(3\theta)$ and $dv = e^{2\theta}d\theta$
Then $du = -3\sin(3\theta)d\theta$ and $v=\frac{1}{2}e^{2\theta}$ 
\begin{align*}
\int e^{2\theta}\cos(3\theta) &= \frac{1}{2}e^{2\theta}\cos(3\theta)-\int (\frac{1}{2}e^{2\theta}\cdot-3\sin(3\theta))d\theta\\
&=\frac{1}{2}e^{2\theta}\cos(3\theta)+ \frac{3}{2} \int e^{2\theta}\sin(3\theta)d\theta
\end{align*}
So you can see I just keep going in circles. How can I break out of this loop?
 [/INST] As lab bhattacharjee answered, in case integration by parts is not mandatory, you can make life easier considering that what you need is the imaginary part of 
$$I=\int e^{2\theta} e^{3i \theta}\,d\theta=\int e^{(2+3i)\theta}\,d\theta=\frac {e^{(2+3i)\theta}}{(2+3i)}=\frac{2-3i}{13}e^{(2+3i)\theta}$$
$$I=\frac{3}{13} e^{2 \theta } \sin (3 \theta )+\frac{2}{13} e^{2 \theta } \cos (3
   \theta )+i \left(\frac{2}{13} e^{2 \theta } \sin (3 \theta )-\frac{3}{13} e^{2
   \theta } \cos (3 \theta )\right)$$
</s>