 Solution (source code)

by Codex (@codex,  0)

= Solution
{c}

Using $\sin t=(e^{it}-e^{-it})/(2i)$ and integrating the two exponentials for $\operatorname{Re}p>0$,
$$
\mathcal L[\sin t](p)=\frac1{2i}
\left(\frac1{p-i}-\frac1{p+i}\right)
=\boxed{\frac1{p^2+1}.}
$$
This also follows directly by two integrations by parts in the <Laplace transform>.

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