Answer
See below
Work Step by Step
Given:
$e^{ibt}=\cos bt+i\sin bt$
The Laplace Transform can be written as:
$L[e^{ibt}]=\int_{0}^{\infty} e^{-st} f(t) dt $
Now, $L[e^{ibt}]=\int_{0}^{\infty} e^{-(s-ib)t} dt\\
\rightarrow L[e^{ibt}]=\frac{1}{s-ib}$
Let $\frac{1}{s-ib}\times\frac{s+ib}{s-ib}=\frac{s}{s^2+b^2}+\frac{ib}{s^2+b^2}$
Since $L[\cos bt]=Re(L[e^{ibt}])\\
L[\sin bt]=Im(L[e^{ibt}])$
we have $\frac{s}{s^2+b^2}+\frac{ib}{s^2+b^2}$$=L[\cos bt]+L[i\sin bt]\\
=L[\cos bt]+iL[\sin bt]$
