Laplace transform of product of $\sinh(t)$ and $\cos(t)$ If I have a function $f(t)=\sinh(t)\cos(t)$ how would I go about finding the Laplace transform? I tried putting it into the integral defining Laplace transformation:
$$
F(s)= \int_0^\infty \mathrm{e}^{-st}\sinh(t)\cos(t)\,\mathrm{d}t
$$
But this integral looks very hairy to me. Can $\sinh(t)\cos(t)$ be rewritten as something more manageable perhaps? 

= L [ { (( e^(-t) - e^(t) ))/2 }.{ cost } ]
Taking Constant outside and separating Laplace to different terms,
= (1/2).{ L[ e^(-t).cost ] - L[ e^(t).cost ] }
Separating solving,
L[ e^(-t).cost ],
first solve ---->  L[ cost ] = s/ (s^(2) + 1)
Therefore for L[ e^(-t).cost ],
Using shifting property,
L[ e^(-t).cost ] = (s+1)/( (s+1)^2 + 1 )  ...........(1)
similarly,
L[ e^(t).cost ] = (s-1)/( (s-1)^2 + 1 )   ............(2)
Subtracting (1) by (2),
will be Answer.
