Answer
$=\frac{s-3}{(s-3)^2+25}-\frac{2}{(s+1)^2+4}$
Work Step by Step
Given: $f(t)=e^{3t}\cos 5t-e^{-t}\sin 2t$
Using the Convolution Theorem
$L[F(s)]=L[e^{3t}\cos 5t-e^{-t}\sin 2t]\\
=L[e^{3t}\cos 5t]-L[e^{-t}\sin 2t]\\
=\frac{s-3}{(s-3)^2+25}-\frac{2}{(s+1)^2+4}$
