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?
HINT:
$$\sinh t\cos t=\frac{e^t-e^{-t}}2\cdot\frac{e^{it}+e^{-it}}2=\frac{e^{t(1+i)}+e^{t(1-i)}-e^{t(i-1)}-e^{-t(1+i)}}4$$

Now, $$L\{e^{at}\}=\frac1{s-a}$$
= 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.