Answer
$L[e^{-4t} \sin 5t]=\dfrac{5}{(s+4)^2+25}$
Work Step by Step
The Laplace transform of function $\sin (5t)$ is given as:
$L(\sin 5t)=\dfrac{s}{s^2+25}$
The first shifting Theorem for $a=-4$ can be expressed as:
$L[e^{-4t} \sin 5t]=\dfrac{5}{(s+4)^2+25}$
