Find the Laplace transform of: $f(t)=\begin{Bmatrix} \sin (\pi t), & 1<t<2\\ 0 & t<1\text{ or }t>2 \end{Bmatrix}$
I know how to find the transform by integration of $\sin( \pi t)$ but I'd like to know how to find the transform by using timeshifting (2. shift).
I know the result is: $F(s)=(e^{-s}-e^{-2s})F_{\sin( \pi t)}(s))=- \frac{\pi (e^{-s}-e^{-2s})}{s^2+\pi^2}$
I've tried to write the function using unit step functions ($\sin(\pi t) u(t-1)-\sin(\pi t) (u(t-2) $ ) but it doesn't seems to work.
We know that $L\{u(t-c)\}=e^{-sc}/s$ and $L\{u(t-c)f(x-c)\}=e^{-sc}L\{f(t)\}$. Hence $$ L\{\sin(\pi t) u(t-1)-\sin(\pi t) (u(t-2) \}=e^{-s}L\{\sin(\pi t+\pi)\}-e^{-2s}L\{\sin(\pi t+2\pi)\}\\ =-e^{-s}L\{\sin(\pi t)\}-e^{-2s}L\{\sin(\pi t)\}\\ =\frac{\pi (-e^{-s}-e^{-2s})}{s^2+\pi^2} $$