Answer
See below
Work Step by Step
Since the given function is periodic on $[0,\infty)$ with period $T=\pi$, we have
$L(f)=\frac{1}{1-e^{-\pi s}}\int ^1_0e^{-st}f(t)dt$
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
$L(f)=\frac{1}{1-e^{-\pi s}}[\frac{2}{\pi}\int ^\frac{\pi}{2}_0 t e^{-st} dt+\int^\pi_{\frac{\pi}{2}}e^{-st} \sin t dt]\\
=\frac{1}{1-e^{-\pi s}}\{[\frac{2}{\pi}(\frac{t e^{-st} }{-s}|^\frac{\pi}{2}_0)-\int^\frac{\pi}{2}_0 \frac{e^{-st}}{-s}dt ]+[\frac{e^{-st}(-s \sin t -\cos t)}{s^2+1}|^\pi_\frac{\pi}{2}]\\
=\frac{1}{1-e^{-\pi s}}[\frac{2-e^{-\frac{s \pi}{2}}(s\pi +2)}{\pi s^2}+\frac{se^{-\frac{s\pi }{2}}+e^{-\pi s}}{s^2+1}]$
