Answer
See below
Work Step by Step
Since the given function is periodic on $[0,\infty)$ with period $T=\pi$, we have
$L(f)=\frac{1}{1-e^{-\pi s}}\int ^\pi_0e^{-st}f(t)dt$
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
$L(f)=\frac{1}{1-e^{-\pi s}}[\int ^{\frac{\pi}{2}}_0 e^{-st} \cos tdt-\int^\pi_{\frac{\pi}{2}}e^{-st}\cos t dt]\\
=\frac{1}{1-e^{-\pi s}}[(\frac{e^{-st} }{-s}|^1_0)- (\frac{e^{-st}}{s}dt |^2_1)]\\
=\frac{1}{1-e^{-\pi s}}[\frac{e^{-\frac{\pi s}{2}}+s}{1+s^2}-\frac{e^{-\pi s}(e^{\frac{\pi s}{2}}-s)}{s^2+1}]\\
=\frac{s(1-e^{-\pi s})+(s+1)e^{-\frac{\pi s}{2}}}{s^2+1}$
