Laplace Question $f(t) = e^{-t} \sin(t)$ I need help with this Laplace question.
$$f(t) = e^{-t} \sin(t) $$

Answer should be $\dfrac{1}{s^2 + 2s + 2}$

What I'm currently doing is as follows:
$u = \sin(t)\qquad$ $dv = e^{-(s+1)t}dt$
$du = \cos(t)dt\qquad$ $v = \dfrac{e^{-(s+1)t}}{-(s+1)}$
$\dfrac{-\sin(t) e^{-(s+1)t}}{-(s+1)}  - \int\dfrac{ e^{-(s+1)t}\cos(t)}{ -(s+1)} dt$
But even if I solved the integral, I wouldn't get this (which is what I should, see picture).



 A: You need not integrate by parts to evaluate this integral.  In fact, one would need to integrate by parts twice.  See the section following the highlighted SPOILER ALERT
So, I thought it would be instructive to present a "trick" that we can use to quickly evaluate the integral of interest and other similar integrals.
HERE IS A HINT:
$$\sin(t)=\text{Im}\left(e^{it}\right) \tag 1$$
SPOILER ALERT:  Scroll over the highlighted area to reveal the solution

Using $(1)$, we have$$\begin{align}\int_0^\infty e^{-t}\sin(t)e^{-st}\,dt&=\text{Im}\left(\int_0^\infty e^{(-1-s+i)t}\,dt\right)\\\\&=\text{Im}\left(\frac{1}{-(1+s)+i)}\right)\\\\&=\text{Im}\left(\frac{-(s+1)-i}{(s+1)^2+1}\,dt\right)\\\\&=\frac{1}{s^2+2s+2}\end{align}$$as was to be shown!  Quick, painless, and less prone to careless errors.


The last result in the OP was
$$\int_0^\infty e^{-t}\sin(t)e^{-st}\,dt=\left.\left(\frac{-\sin(t) e^{-(s+1)t}}{-(s+1)}\right)\right|_0^\infty  - \int_0^{\infty}\dfrac{ e^{-(s+1)t}\cos(t)}{ -(s+1)} dt$$
Continuing we have
$$\begin{align}
\int_0^\infty e^{-t}\sin(t)e^{-st}\,dt&=\frac{1}{s+1} \int_0^{\infty} e^{-(s+1)t}\cos(t)\,dt \tag 2
\end{align}$$
We integrate by parts $(2)$ with $u=\cos(t)$ and $v=-\frac{e^{-(s+1)t}}{s+1}$.  Then, we obtain
$$\begin{align}
\int_0^\infty e^{-t}\sin(t)e^{-st}\,dt&=\frac{1}{s+1}\left.\left(-\frac{\cos(t)e^{-(s+1)t}}{s+1}\right)\right|_0^\infty -\frac{1}{(s+1)^2}\int_0^\infty \sin(t)e^{-(s+1)t}\,dt\\\\
&=\frac{1}{(s+1)^2}-\frac{1}{(s+1)^2}\int_0^\infty \sin(t)e^{-(s+1)t}\,dt\\\\
((s+1)^2+1)\int_0^\infty e^{-t}\sin(t)e^{-st}\,dt&=1\\\\
\int_0^\infty e^{-t}\sin(t)e^{-st}\,dt&=\frac{1}{s^2+2s+2}
\end{align}$$
as expected!
A: Here's the 'direct' proof:
$f(t)=e^{-t}\sin t.$
\begin{eqnarray*}\mathcal{L}\{f(t)\}&=&\int_{0}^{+\infty}e^{-(s+1)t}\sin t\,dt~=~\lim_{\ell\to+\infty}\int_{0}^{\ell}e^{-(s+1)t}\sin t\,dt\\
   &=&-\frac{1}{s+1}\lim_{\ell\to+\infty}\left[e^{-(s+1)t}\sin t\Big|_{0}^{\ell}-\int_{0}^{\ell}e^{-(s+1)t}\cos t\,dt\right]\\
   &=&\frac{1}{s+1}\lim_{\ell\to+\infty}\int_{0}^{\ell}e^{-(s+1)t}\cos t\,dt,
 \end{eqnarray*}
since $$\lim_{\ell\to+\infty}e^{-(s+1)\ell}\sin\ell=0.$$ Integrating again by parts yields
\begin{eqnarray*}\mathcal{L}\{f(t)\}&=&-\frac{1}{(s+1)^2}\lim_{\ell\to+\infty}\left[e^{-(s+1)t}\cos t\Big|_{0}^{\ell}+\int_{0}^{\ell}e^{-(s+1)t}\sin t\,dt\right]\\
  &=&\frac{1}{(s+1)^2}\left[1-\lim_{\ell\to+\infty}\int_{0}^{\ell}e^{-(s+1)t}\sin t\,dt\right]\\
  &=&\frac{1}{(s+1)^2}(1-\mathcal{L}\{f(t)\})
 \end{eqnarray*}
and finally solving by $\mathcal{L}\{f(t)\}$,
$$\boxed{\mathcal{L}\{f(t)\}=\frac{1}{(s+1)^2+1}=\frac{1}{s^2+2s+2}}~.$$
A: Calculate Laplace transform for $f(t)$ and $e^{-t}\cos(t)$ then you'll get a equation system where the unknowns will be the integrals 
$$\int\dfrac{ e^{-(s+1)t}\cos(t)}{ -(s+1)} dt \text{ and } \int\dfrac{ e^{-(s+1)t}\sin(t)}{ -(s+1)} dt $$
with this you can conclude.
