Solution
= Solution
For one period,
$$
\int_0^{2\pi}e^{-st}h(t)dt=\int_0^\pi e^{-st}\sin t\,dt
=\frac{1+e^{-\pi s}}{s^2+1}.
$$
Part (b), followed by cancellation of $1+e^{-\pi s}$, gives
$$
\boxed{\mathcal L\{h\}(s)=\frac{1}{(s^2+1)(1-e^{-\pi s})}.}
$$