Find $\int e^{2\theta} \cdot \sin{3\theta} \ d\theta$ I am working on an integration by parts problem that, compared to the student solutions manual, my answer is pretty close. Could someone please point out where I went wrong?

Find $\int e^{2\theta} \cdot \sin{3\theta} \ d\theta$

$u_1 = \sin{3\theta}$
$du_1 = \frac{1}{3}\cos{3\theta} \ d\theta$
$v_1 = \frac{1}{2} e^{2\theta}$
$dv_1 = e^{2\theta} \ d\theta$

$\underbrace{\sin{3\theta}}_{u_1} \cdot \underbrace{\frac{1}{2} e^{2\theta}}_{v_1} - \int\underbrace{\frac{1}{2} e^{2\theta}}_{v_1} \cdot \underbrace{\frac{1}{3} \cos{3\theta} \ d\theta}_{du_1}$
$\frac{1}{2}\sin{3\theta} \cdot e^{2\theta} - \frac{1}{6} \int \cos{3\theta} \cdot e^{2\theta} \ d\theta$

Doing integration by parts again...
$u_2 = \cos{3\theta}$
$du_2 = -\frac{1}{3} \sin{3\theta} \ d\theta$
$v_2 = \frac{1}{2} e^{2\theta}$
$dv_2 = e^{2\theta} \ d\theta$

$\underbrace{\sin{3\theta}}_{u_1} \cdot \underbrace{\frac{1}{2} e^{2\theta}}_{v_1} - \frac{1}{6}\left(\underbrace{\cos{3\theta}}_{u_2} \cdot \underbrace{\frac{1}{2} e^{2\theta}}_{v_2} - \int \underbrace{\frac{1}{2} e^{2\theta}}_{v_2} \cdot \underbrace{-\frac{1}{3} \sin{3\theta} \ d\theta}_{du_2}\right)$
$\frac{1}{2}\sin{3\theta} \cdot e^{2\theta} - \frac{1}{6}\left(\frac{1}{2}\cos{3\theta} \cdot e^{2\theta} + \frac{1}{6} \int \sin{3\theta} \cdot e^{2\theta} \ d\theta\right)$
$\frac{1}{2}\sin{3\theta} \cdot e^{2\theta} - \frac{1}{12}\cos{3\theta} \cdot e^{2\theta} - \frac{1}{36} \int \sin{3\theta} \cdot e^{2\theta} \ d\theta$
$\frac{37}{36} \int \sin{3\theta} \cdot e^{2\theta} \ d\theta = \frac{1}{2}\sin{3\theta} \cdot e^{2\theta} - \frac{1}{12}\cos{3\theta} \cdot e^{2\theta}$
$\int \sin{3\theta} \cdot e^{2\theta} \ d\theta = \begin{equation} \boxed{\frac{18}{37}\sin{3\theta} \cdot e^{2\theta} - \frac{1}{37}\cos{3\theta} \cdot e^{2\theta}} \end{equation}$

However, the boxed answer is incorrect. The answer should read:

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

 A: Ah ha! I got it!

$du = 3 \cos{3\theta} \ d\theta$ not $\frac{1}{3} \cos{3\theta} \ d\theta$

For some reason I must have been thinking of integrating the dv rather than differentiating.
A: One could also use the right tool, which is the following fact:

For every nonzero complex number $z$, a primitive of $\theta\mapsto\mathrm e^{z\theta}$ is $\theta\mapsto z^{-1}\mathrm e^{z\theta}$.

Here, one is looking at the imaginary part of $\mathrm e^{z\theta}$ with $z=2+3\mathrm i$, hence the answer is the imaginary part of 
$$
\theta\mapsto z^{-1}\mathrm e^{z\theta}=|z|^{-2}\bar z\mathrm e^{z\theta}=|z|^{-2}(2-3\mathrm i)\mathrm e^{2\theta}(\cos(3\theta)+\mathrm i\sin(3\theta)).
$$
Since $|z|^2=2^2+3^2=13$ and the imaginary part of $(2-3\mathrm i)(\cos(3\theta)+\mathrm i\sin(3\theta))$ is $2\sin(3\theta)-3\cos(3\theta)$, an answer is $\theta\mapsto \frac1{13}\mathrm e^{2\theta}(2\sin(3\theta)-3\cos(3\theta))$.
A: $$\int \mathrm{e}^{ax}\sin(bx)dx
=\frac{\mathrm{e}^{ax}}{\mathrm{a^2+b^2}}[a\sin(bx)-b\cos(bx)]+c$$
