Evaluate $\int e^{2\theta} \sin (3\theta)\ d\theta$ 
Evaluate $$\int e^{2\theta} \sin (3\theta)\ d\theta .$$

I am little stuck as to what I can do after this point. Please tell me if my method overall is flawed:

 A: An alternative:
Work in reverse and try differentiating the expression "to see",
$$(e^{2t}\sin(3t))'=2e^{2t}\sin(3t)+3e^{2t}\cos(3t).$$
There is a similar new term, with a cosine. Have a look at its derivative,
$$(e^{2t}\cos(3t))'=2e^{2t}\cos(3t)-3e^{2t}\sin(3t).$$
Then, forming a suitable linear combination to eliminate the cosine, you get
$$(2e^{2t}\sin(3t)-3e^{2t}\cos(3t))'=13e^{2t}\sin(3t).$$

More generally, $e^{ax}\sin/\cos(bx)$ will yield
$$(e^{ax}(a\sin(bx)-b\cos(bx)))'=(a^2+b^2)e^{ax}\sin(bx),\\
(e^{ax}(a\cos(bx)+b\sin(bx)))'=(a^2+b^2)e^{ax}\cos(bx).$$
A: If you denote $I := \int e^{2 \theta} \sin 3 \theta$, then your computation using integration by parts twice (which is the standard approach, and a reasonable one) says that
$$I = f(\theta) + \frac{9}{4} I$$
for some function $f$ that does not involve an integral. (Perhaps there is a sign error in your formula, but this does not affect the applicability of our method:) Hence by algebra we can simply solve for $I$ (i.e., evaluate the integral) in terms of $f$. (Don't forget to include the general constant, i.e, the "$+ C$", which should appear in the formula for your general antiderivative but does not appear in the above equality because, as usual, we interpret $I$ to be a family of functions, all equal up to an overall constant.)
If you're comfortable with complex numbers, one can also evaluate the integral by recalling that $\sin \alpha = \frac{1}{2 i}(e^{i \alpha} - e^{-i \alpha})$ distributing, using that $\int e^{\beta t} dt = \frac{1}{b} e^{\beta t} + C$, and rewriting your expression in real terms.
A: You  have some mistake:
$$
\int e^{2\theta}\sin(3\theta) d\theta=\frac{1}{2}e^{2\theta}\sin(3\theta)-\frac{3}{2}\int e^{2\theta}\cos(3\theta) d\theta=
$$
$$
=\frac{1}{2}e^{2\theta}\sin(3\theta)-\frac{3}{4} e^{2\theta}\cos(3\theta) -\frac{9}{4}\int e^{2\theta}\sin(3\theta) d\theta
$$
so:
$$
\left(1+\frac{9}{4} \right)\int e^{2\theta}\sin(3\theta) d\theta=\frac{1}{2}e^{2\theta}\sin(3\theta)-\frac{3}{4} e^{2\theta}\cos(3\theta)
$$
and
$$
\int e^{2\theta}\sin(3\theta) d\theta=\frac{e^{2\theta}}{13}\left(2\sin(3\theta)-3\cos(3\theta)\right)
$$
A: It is much simpler to compute with  the complex exponential function: $\;\mathrm e^{2\theta}\sin 3\theta$ is the imaginary part of $\;\mathrm e^{(2+3\mathrm i)\theta}$. So we compute:
\begin{align*}
\int \mathrm e^{(2+3\mathrm i)\theta}\,\mathrm d\mkern1mu\theta&=\frac1{2+3\mathrm i} \mathrm e^{(2+3\mathrm i)\theta}=\frac{2-3\mathrm i}{13} \mathrm e^{(2+3\mathrm i)\theta}\\
&=\frac1{13}\mathrm e^{2\theta}\bigl(2\cos 3\theta+3\sin3\theta+\mathrm i(2\sin3\theta-3\cos 3\theta)\bigr),
\end{align*}
and take the imaginary part of the result:
$$\int\mathrm e^{2\theta}\sin 3\theta\,\mathrm d\mkern1mu\theta=\frac1{13}\mathrm e^{2\theta}(2\sin3\theta-3\cos 3\theta).$$
A: Observe
$$(e^{at}\sin(bt+\phi))'=e^{at}(a\sin(bt+\phi)+b\cos(bt+\phi))=e^{at}r\sin(bt+\phi+\delta),$$
where $r=\sqrt{a^2+b^2}$ and $\tan(\delta)=\dfrac ba$.
The derivative of a damped sinusoid is that damped sinusoid with a change in amplitude and in phase.
It suffices to reverse these changes to get
$$\int e^{at}\sin(bt)dt=e^{at}\frac 1r\sin(bt-\delta)=e^{at}\frac 1r(\sin(bt)\frac ar-\cos(bt)\frac br).$$
