Answer
See below
Work Step by Step
Since the given function is periodic on $[0,\infty)$ with period $T=2a$, we have
$L(f)=\frac{1}{1-e^{-2sa}}\int ^{2a}_0e^{-st}f(t)dt$
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
$L(f)=\frac{1}{1-e^{-2sa}}[\frac{1}{a}\int ^{a}_0 t e^{-st} \cos tdt+\frac{1}{a}\int^{2a}_{a}(2a-t)e^{-st}dt]\\
=\frac{1}{1-e^{-2sa}}[\frac{e^{-st}(-1-st) }{as^2}|^a_0+(\frac{e^{-2as+su}(su-1)}{as^2} |^{2a}_a)]$
With $u=2a-t$
$=\frac{1}{1-e^{-2sa}}\{\frac{1-e^{-sa}(1+sa)}{as^2}+\frac{e^{-sa}[1+e^{sa}(sa-1)]}{as^2}\}\\
=\frac{1}{1-e^{-2sa}}[\frac{1-e^{-sa}(1+sa)}{as^2}]$
