I am puzzled by this. If you learned "how to take a Laplace Transform" why do you not know how to find the Laplace transform of cos(at)? The Laplace transform of a function, f(x), is \(\displaystyle \int_0^\infty f(t)e^{-st}dt\). In particular applying that to \(\displaystyle cos(at)\) means finding \(\displaystyle \int_0^\infty cos(at)e^{st}dt\). Since you are integrating a product of two functions, that typically means "integration by parts". Here, let \(\displaystyle u(t)= cos(at)\) and \(\displaystyle dv= e^{-st}dt\). Then \(\displaystyle du= -a sin(at)dt\) and \(\displaystyle v= -\frac{1}{s}e^{-st}\). So \(\displaystyle \int_0^\infty cos(ax)e^{-st} dt= \left[\frac{1}{s}cos(at)e^{-st}\right]_0^\infty- \frac{a}{s}\int_0^\infty sin(at)e^{-st}dt= -\frac{1}{s}- \frac{a}{s}\int_0^\infty sin(ax)e^{-st}dt\). Now, use integration by parts again. Let u= sin(at) and let \(\displaystyle dv= e^{-st}dt\). Then \(\displaystyle du= acos(at)dt\) and \(\displaystyle v= -\frac{1}{s}e^{-st}dt\). \(\displaystyle \int_0^\infty sin(at)e^{-st}dt= \left[-\frac{1}{s}sin(at)e^{-st}\right]_0^\infty+ \frac{a}{s}\int_0^\infty cos(at)e^{-st}dt= \frac{a}{s}\int_0^\infty cos(at)e^{-st}dt\).
Putting that into the first integral, \(\displaystyle \int_0^\infty cos(t)e^{-st}dt= \frac{1}{s}- \frac{a^2}{s^2}\int_0^\infty cos(t)e^{-st}dt\). Then \(\displaystyle \left(1+ \frac{a^2}{s^2}\right)\int_0^\infty cos(t)e^{-st}dt= \frac{1}{s}\). Dividing both sides by \(\displaystyle 1+ \frac{a^2}{s^2}=\frac{s^2+ a^2}{s^2}\) gives \(\displaystyle \int_0^\infty cos(t)e^{-st}dt= \frac{s}{s^2+ a^2}\).
It is important for differential equations because the Laplace transform of the derivative of function f is an algebraic formula: To find \(\displaystyle \int_0^\infty f'(t)e^{-st}dt\) again use integration by parts letting \(\displaystyle u= e^{-st}\) and \(\displaystyle dv= f'(t)dt\) so that \(\displaystyle du= -se^{-st}dt\) and \(\displaystyle v= f(t)\). So \(\displaystyle \int_0^\infty f'(t)e^{-st}dt= \left[f(t)e^{-st}\right]_0^\infty- s\int_0^\infty f(t)e^{-st}dt= f(0)- s\int_0^\infty f(t)e^{-st}dt\). That is, the Laplace transform of the derivative of f is f(0) minus the Laplace transform of f. A differential equation in f can be converted to an algebraic equation for the Laplace transform of f.
