Laplace transform of $te^{-2t}\sin(2t)u(t-3)$
I do know the following properties of Laplace Transform:
A) $t f(t) = \frac {dF(S)}{ds}$
B) $e^{at} f(t) = F(S+a)$
But from what I see a part of the function is in terms of $t$: $te^{-2t}\sin(2t)$ and the other in terms of $t-3$: $u(t-3)$
So how do I approach this?
Also this is the waveform of $u(t-3)$
Hint: Take $f(t)=t\sin(2t)$, and rewrite it as $\Im(te^{i2t})$, find $\Im(\mathscr{L}\{f(t)\})$ by applying $s$-shifting once. To the result of this, apply $s$-shifting again with $\mathscr{L}\{e^{-4t}f(t)\}$. On getting this, multiply the result by $e^{-3s}$, which follows from the $t$-shifting theorem.
If you wish to use the definition, you may write the following and apply integration by parts to arrive at the answer.
$$\mathscr{L}\{te^{-4t}\sin(2t)u(t-3)\}=\int_{3}^{\infty}te^{-4t}\sin(2t)dt$$