Any special way to integrate $\int_{0}^{\infty}t^2\exp(-t)\sin(t)$ quickly? Is there some trick to quickly integrate

$$\int_{0}^{\infty}t^2\exp(-t)\sin(t)$$ 

or is there a way one can see that this integral equals $\frac{1}{2}$?
I'm asking because I am preparing for a test and it seems like integration by parts would take far too long.
 A: Since $\int_0^\infty t^2 e^{-zt}dt=\frac{2}{z^3}$ if $\Re z>0$, the desired integral is $$\Im\int_0^\infty t^2\exp\Big[-(1-i)t\Big]dt=\Im\frac{2}{(1-i)^3}=\Im\frac{1}{\sqrt{2}}e^{3\pi i/4}=\frac{1}{\sqrt{2}}\sin\frac{3\pi}{4}=\frac{1}{2}.$$
A: There are other methods to be "quick" in certain calculations. In this case one may start with 
$$\int_{0}^{\infty} e^{-st} \, \sin(at) \, dt = \frac{a}{s^2 + a^2}.$$
Since $$D_{a}^{2} \, \sin(a t) = - t^{2} \, \sin(at)$$ 
and 
$$ D_{s}^{2} \, e^{-s t} = t^{2} \, e^{-st}$$
then:
\begin{align}
\int_{0}^{\infty} t^{2} \, e^{-t} \, \sin(at) \, dt &= - D_{a}^{2} \, \int_{0}^{\infty} e^{- t} \, \sin(a t) \, dt \\
&= - D_{a}^{2} \left( \frac{a}{a^2 + 1}\right) \\
&= \frac{2 a \, (3 - a^2)}{(a^2 + 1)^{3}}
\end{align}
or 
\begin{align}
\int_{0}^{\infty} t^{2} \, e^{-s t} \, \sin(t) \, dt &= D_{s}^{2} \, \int_{0}^{\infty} e^{-s t} \, \sin(t) \, dt \\
&= D_{s}^{2} \left(\frac{1}{s^2 + 1}\right) \\
&= \frac{2 (3 s^2 - 1)}{(s^2 + 1)^{3}}.
\end{align}
In these cases if $a=1$ or $s=1$ the resulting values lead to
$$ \int_{0}^{\infty} t^{2} \, e^{-t} \, \sin(t) \, dt = \frac{1}{2}. $$
