Laplace Transform gets complicated

Question

The Laplace transform of $\cos(at)$ defined as

$\mathcal{L}\{\cos(at)\} = \int_{0}^{\infty}dt ~e^{-st} \cos(at) = \frac{s}{s^2 + a^2}~~~~~~~~s>0$

How to find $\int_{0}^{\infty}dt ~e^{-st} \cos[\Big(\sqrt{(\frac{a}{2s})^2 -1}\Big)st]$, or, $~\mathcal{L}\{\cos[\Big(\sqrt{(\frac{a}{2s})^2 -1}\Big)st]\}$?

Answer 1

$$\int_{0}^{\infty} ~e^{-st} \cos[\Big(\sqrt{(\frac{a}{2s})^2 -1}\Big)st]\ dt$$

Let $\Big(\sqrt{(\frac{a}{2s})^2 -1}\Big)s=k(const.)$ since it doesn't have $t$ term.

$$\implies \int_{0}^{\infty} ~e^{-st} \cos(kt)dt = \frac{s}{s^2 + k^2}~~~~~~~~s>0$$