How to compute $ \int e^{-st} \sin(2t) dt $ Wolfram Alpha shows me the result of $ \int  e^{-st} \sin(2t) dt $ .
However it doesn't let me see the step to step solution. Then I tried to do this by hand as the solution didn't look "too difficult", however couldn't do it. 
So can someone show me how to do it? 
 A: You can integrate by parts twice or use this method
$$\int e^{-st}\sin (2t)dt=\operatorname{Im}\int e^{(-s+2i)t}dt$$
Can you take it from here?
A: using integration by parts we obtain
$\int e^{-st}\sin(2t)dt=-e^{-st}\frac{cos(2t)}{2}-\int\frac{s}{2}e^{-st}\cos(2t)dt$
and now the same once more
A: You start by using integration by parts two times, and then solve algebraically for the integral you want:
$\begin{align}
\int  e^{-st} \sin(2t) dt  &= -\frac12 e^{-st}\cos(2t) -\frac{s}{2}\int e^{-st}\cos(2t) dt \\
&=-\frac12 e^{-st}\cos(2t) -\frac{s}{2}\left(\frac12 e^{-st}\sin(2t) + \frac{s}{2}\int e^{-st}\sin(2t) dt \right) \\
&=-\frac12 e^{-st}\cos(2t)  -\frac{s}{4}e^{-st}\sin(2t) - \frac{s^2}{4}\int e^{-st}\sin(2t) dt
\end{align}$
So, that's two applications of integration by parts. Now you can add that last term back over to the left-hand side:
$\int  e^{-st} \sin(2t) dt + \frac{s^2}{4}\int e^{-st}\sin(2t) dt = -\frac12 e^{-st}\cos(2t)  -\frac{s}{4}e^{-st}\sin(2t)$
Multiplying everything by $4$ and factoring appropriately, we obtain:
$(4+s^2)\int e^{-st}\sin(2t) dt = -e^{-st}\left(2\cos(2t) + s\sin(2t)\right)$
or,
$\int e^{-st}\sin(2t) dt = \frac{-e^{-st}\left(2\cos(2t) + s\sin(2t)\right)}{4+s^2}$.
