Find the Laplace transforms of the functions given in problem f(t)=sin pi t text{ if } 2leq tleq3 , f(t)=0 text{ if } t<2 text{ or if } t>3

Step 1
Given function is
\(\begin{cases}\sin \pi t & 2\leq t\leq3\0 & 23\end{cases}\)
The Laplace transform of function f(t) is defined by
$L[f(t)]={\int }_0^{\mathrm{\infty }}{e}^{-st}f(t)dt$
$={\int }_0^2{e}^{-st}f(t)dt+{\int }_2^3{e}^{-st}f(t)dt+{\int }_3^{\mathrm{\infty }}{e}^{-st}f(t)dt$
$={\int }_2^3{e}^{-st}f(t)dt\{\because f(t)=0,2<t\text{ or }t>3\}$
$={\int }_2^3{e}^{-st}\sin (\pi t)dt$
Step 2
Now, we firstly evaluate the indefinite integral by parts.
$I=\int e^{-st}\sin (\pi t)dt=\sin (\pi t)\int e^{-st}dt-\int (\left(\frac{d}{dt}\right)\sin (\pi t)\int e^{-st}dt)dt$
$I=-\frac{e^{-st}}{s}\sin (\pi t)-\int \pi \cos (\pi t)(-\frac{e^{-st}}{s})dt$
$I=-\frac{e^{-st}}{s}\sin (\pi t)+\frac{\pi }{s}\int e^{-st}\cos (\pi t)dt$
Again , we integrate by parts.
$I=-\frac{e^{-st}}{s}\sin (\pi t)+\frac{\pi }{s}[\cos (\pi t)\int e^{-st}dt-\int (\frac{d}{dt}\cos \pi t\int e^{-st}dt)dt]$
$I=-\frac{e^{-st}}{s}\sin (\pi t)+\frac{\pi }{s}[-\frac{e^{-st}}{s}\cos (\pi t)-\int (-\pi \sin (\pi t))(-\frac{e^{-st}}{s})dt]$
$I=-\frac{e^{-st}}{s}\sin (\pi t)+\frac{\pi }{s}[-\frac{e^{-st}}{s}\cos (\pi t)-\frac{\pi }{s}\int e^{-st}\sin (\pi t)dt]$
$I=-\frac{e^{-st}}{s}\sin (\pi t)-\frac{\pi e^{-st}}{s^2}\cos (\pi t)-\frac{{\pi }^2}{s^2}I$
$I+\frac{{\pi }^2}{s^2}I=-\frac{e^{-st}}{s}\sin (\pi t)-\frac{\pi e^{-st}}{s^2}\cos (\pi t)$
$I=-\frac{e^{-st}}{{\pi }^2+s^2}[s(\sin (\pi t))+\pi \cos (\pi t)]$
Step 3
Now, we evaluate the Laplace transform of given function.
$L[f(t)]={[-\frac{e^{-st}}{{\pi }^2+s^2}[s(\sin (\pi t))+\pi \cos (\pi t)]]}_2^3$