Problem 02 | First Shifting Property of Laplace Transform

$\mathcal{L} (\sin bt) = \dfrac{b}{s^2 + b^2}$

$\mathcal{L} (\sin 3t) = \dfrac{3}{s^2 + 3^2}$

$\mathcal{L} (\sin 3t) = \dfrac{3}{s^2 + 9}$
 

Thus,
$\mathcal{L} (e^{-5t}\sin 3t) = \dfrac{3}{(s + 5)^2 + 9}$

$\mathcal{L} (e^{-5t}\sin 3t) = \dfrac{3}{(s^2 + 10s + 25) + 9}$

$\mathcal{L} (e^{-5t}\sin 3t) = \dfrac{3}{s^2 + 10s + 34}$           answer