Derive the Laplace transform of the following function sin(h(at))

Step 1 We have to derive the Laplace transform of function:
$\sin (h(at))$
We know from definition of hyperbolic function,
$\sin hx=\frac{e^x-e^{-x}}{2}$
If x=at then,
$\sin hx=\frac{e^x-e^{-x}}{2}$
$\sin hat=\frac{e^{at}-e^{-at}}{2}$
$=\frac{1}{2}(e^{at}-e^{-at})$
Step 2
Now finding Laplace transform,
$L(\sin hat)=L(\frac{1}{2}(e^{at}-e^{-at})$
$=\frac{1}{2}(L{e}^{at}-L(e^{-at}))$
We know the Laplace transform of exponential function,
$L(e^{ax})=\frac{1}{(s-a)}$
if a=-a then $L(e^{-ax})=\frac{1}{(s+a)}$
Putting Laplace transform of exponential function, we get
$\frac{1}{2}(L{e}^{at}-L(e^{-at}))=\frac{1}{2}((\frac{1}{(s-a)})-(\frac{1}{(s+a)})$
$=\frac{1}{2}\frac{(s+a-(s-a))}{(s-a)(s+a)}$
$=\frac{1}{2}\frac{(s+a-s+a)}{(s^2-a^2)}\text{ since }(\sin ce(a+b)(a-b)=a^2-b^2)$
$=\frac{1}{2}\frac{2a}{(s^2-a^2)}$
$=\frac{a}{s^2-a^2}$
Hence, Laplace transform of $\sin hat$ is $\frac{a}{(s^2-a^2)}$