Answer
$$F\left( s \right) = \frac{a}{{{s^2} + {a^2}}},{\text{ }}s > 0$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( t \right) = \sin at \cr
& {\text{Find the Laplace Transform of the function }} \cr
& F\left( s \right) = \int_0^\infty {{e^{ - st}}f\left( t \right)} dt, \cr
& {\text{Substitute }}f\left( t \right) = \sin at \cr
& F\left( s \right) = \int_0^\infty {{e^{ - st}}\sin at} dt \cr
& {\text{By the Definition of Improper Integrals with Infinite }} \cr
& {\text{Integration Limits}} \cr
& \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx,{\text{ then}} \cr
& F\left( s \right) = \mathop {\lim }\limits_{b \to \infty } \int_0^b {{e^{ - st}}\sin at} dt{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Integrate by tables, using the formula }} \cr
& \int {{e^{mu}}\sin nt} du = \frac{{{e^{mu}}}}{{{m^2} + {n^2}}}\left( {m\sin nu - n\cos nu} \right) + C \cr
& {\text{Then for }}\int {{e^{ - st}}\sin at} {\text{ Let }}m = - s,{\text{ }}n = a,{\text{ }}t = u,{\text{ so}} \cr
& \int {{e^{ - st}}\sin at} = \frac{{{e^{ - st}}}}{{{{\left( { - s} \right)}^2} + {{\left( a \right)}^2}}}\left( { - s\sin at - a\cos at} \right) + C \cr
& \int {{e^{ - st}}\sin at} = \frac{{{e^{ - st}}}}{{{s^2} + {a^2}}}\left( { - s\sin at - a\cos at} \right) + C \cr
& {\text{Substituting the previous result into }}\left( {\bf{1}} \right) \cr
& F\left( s \right) = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{{{e^{ - st}}}}{{{s^2} + {a^2}}}\left( { - s\sin at - a\cos at} \right)} \right]_0^b \cr
& F\left( s \right) = \frac{1}{{{s^2} + {a^2}}}\mathop {\lim }\limits_{b \to \infty } \left[ {{e^{ - st}}\left( { - s\sin at - a\cos at} \right)} \right]_0^b \cr
& F\left( s \right) = \frac{1}{{{s^2} + {a^2}}}\mathop {\lim }\limits_{b \to \infty } \left[ {{e^{ - sb}}\left( { - s\sin bt - a\cos bt} \right)} \right] \cr
& - \frac{1}{{{s^2} + {a^2}}}\mathop {\lim }\limits_{b \to \infty } \left[ {{e^0}\left( { - s\sin 0 - a\cos 0} \right)} \right] \cr
& F\left( s \right) = \frac{1}{{{s^2} + {a^2}}}\mathop {\lim }\limits_{b \to \infty } \left[ {{e^{ - sb}}\left( { - s\sin bt - a\cos bt} \right)} \right] - \frac{1}{{{s^2} + {a^2}}}\mathop {\lim }\limits_{b \to \infty } \left[ { - a} \right] \cr
& {\text{Evaluate the limit when }}b \to \infty ,{\text{ }}s > 0 \cr
& F\left( s \right) = \frac{1}{{{s^2} + {a^2}}}\left[ {{e^{ - \infty }}\left( { - s\sin \left( \infty \right) - a\cos \left( \infty \right)} \right)} \right] - \frac{1}{{{s^2} + {a^2}}}\left[ { - a} \right] \cr
& F\left( s \right) = \frac{1}{{{s^2} + {a^2}}}\left[ 0 \right] - \frac{1}{{{s^2} + {a^2}}}\left[ { - a} \right] \cr
& F\left( s \right) = \frac{a}{{{s^2} + {a^2}}},{\text{ }}s > 0 \cr} $$
