Answer
See below
Work Step by Step
Since the given function is periodic on $[0,∞)$ with period $T=\pi$, we have
$L(f)=\frac{1}{1-e^{-\pi s}}\int ^\pi_0e^{-st}f(t)dt=\frac{1}{1-e^{-\pi s}}\int ^\pi_0 \sin t e^{-st} dt$
Using the standard integral
$\int e^{at}\sin bt dt=\frac{1}{a^2+b^2}e^{at}(a\sin bt-b\cos bt)+c$
it follows that
$L(f)=\frac{1}{1-e^{-\pi s}}\int ^\pi_0\sin t e^{-st} dt\\
=\frac{1}{1-e^{-\pi s}}[\frac{e^{-st}(-s\cos t -\sin t)}{s^2+1}]^\pi_0\\
=\frac{1+e^{-\pi s}}{1-e^{-\pi s}}[\frac{s}{s^2+1}]$
