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
∫0∞f(t)e−stdt. In particular applying that to
cos(at) means finding
∫0∞cos(at)estdt. Since you are integrating a product of two functions, that typically means "integration by parts". Here, let
u(t)=cos(at) and
dv=e−stdt. Then
du=−asin(at)dt and
v=−s1e−st. So
∫0∞cos(ax)e−stdt=[s1cos(at)e−st]0∞−sa∫0∞sin(at)e−stdt=−s1−sa∫0∞sin(ax)e−stdt. Now, use integration by parts again. Let u= sin(at) and let
dv=e−stdt. Then
du=acos(at)dt and
v=−s1e−stdt.
∫0∞sin(at)e−stdt=[−s1sin(at)e−st]0∞+sa∫0∞cos(at)e−stdt=sa∫0∞cos(at)e−stdt.
Putting that into the first integral,
∫0∞cos(t)e−stdt=s1−s2a2∫0∞cos(t)e−stdt. Then
(1+s2a2)∫0∞cos(t)e−stdt=s1. Dividing both sides by
1+s2a2=s2s2+a2 gives
∫0∞cos(t)e−stdt=s2+a2s.
It is important for differential equations because the Laplace transform of the
derivative of function f is an algebraic formula: To find
∫0∞f′(t)e−stdt again use integration by parts letting
u=e−st and
dv=f′(t)dt so that
du=−se−stdt and
v=f(t). So
∫0∞f′(t)e−stdt=[f(t)e−st]0∞−s∫0∞f(t)e−stdt=f(0)−s∫0∞f(t)e−stdt. 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.