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