Laplace transform of multiplication of three terms Okay, so I have 
$${f}(t)= t\mathrm{e}^{-2t}\sin 2t.$$
In order to do a Laplace transform, I'm pretty positive I cannot just split it up cause that would basically break the rules of math. I understand how to do a transform with just two, not three, t's. 
Like I know that 
\begin{eqnarray*}
\mathcal{L}\left[t\mathrm{e}^{-2t}\right](s) &=& \frac{1}{(s+2)^2} \\ \\ \\  \\
\mathcal{L}\left[\mathrm{e}^{-2t}\sin 2t \right](s) &=& \frac{2}{(s+2)^2+4}
\end{eqnarray*}
But how do I find $\mathcal{L}\left[\mathrm{f}(t)\right](s)$?
Help!
 A: Hint: Use the identity

$$ \sin 2t = \frac{1}{2i}\left( e^{i2t} - e^{-i2t}\right), $$

and the linearity of Laplace transform.
Added: You can go this way, first you can use the Laplace transform of $\sin 2t$ 

$$ F(s) = \frac{2}{s^2+4}, $$

then use the two properties 

$$ e^{at}=F(s-a), $$

and

$$ t^nf(t) = (-1)^n F^{(n)}(s) $$

in a row.
A: You are almost there! Just let me get my notation clear.
Let $\mathrm{f}$ be a suitable function, and define the Laplace Transform of $\mathrm{f}$ as $\mathrm{F}$, given by
$$\mathrm{F}(s) = \int_0^{\infty} \mathrm{f}(t)~\mathrm{e}^{-st}~\mathrm{d}t \ \ \text{where } \ \ \ s>0$$
If $\mathrm{f}(t) = \mathrm{e}^{-2t}\sin 2t$ then you have already shown that
$$\mathrm{F}(s) = \frac{2}{(s+2)^2+4}$$
There is one -- among many -- nice little facts about the Laplace Transform. One of them is that the Laplace Transform of $t\,\mathrm{f}(t)$ is given by the negative of the, i.e. derivative $\displaystyle{-\frac{\mathrm{dF}}{\mathrm{d}s}}$, where $\mathrm{F}$ is the Laplace Transform of $\mathrm{f}$.
In other words, the Laplace Transform of $t\,\mathrm{e}^{-2t}\sin 2t$ is given by
$$-\frac{\mathrm{d}}{\mathrm{d}s}\left(\frac{2}{(s+2)^2+4}\right)$$
Can you find this derivative using, for example, the quotient rule?
