Best Answer:
F(s) = integral from t = 0 to infinity of

[cos(at)] * e^(-st) dt

Integrate by parts, letting

u = cos(at) and dv = e^(-st) dt

Then

du = -a sin(at) dt and v = (-1/s) e^(-st)

The integral is

uv - integral of v du

= [cos(at) * (-1/s)e^(-st)] +

integral of (-a/s) [sin(at)*e^(-st)] dt

The first term, evaluated at t = infinity and zero,

comes out to just 1/s.

The second term must be integrated by parts a second time,

using u2 = a sin(at) and dv2 = (-1/s)e^(-st) dt, so

du2 = a^2 cos(at) and v2 = (-1/s^2)e^(-st)

Then the original integral must be

1/s - (a/s^2)*sin(at)e^(-st) -

(a^2/s^2) integral of [ (cos at) e^(-st) ] dt

ADD the integral from the RHS to the original integral on the LHS, obtaining:

[1 + (a^2/s^2)] * F(s) =

1/s + [(a/s^2)*(sin at)*e^(-st)]

but note that the last term will be zero BOTH at t = infinity and t = 0.

So you have

F(s) = (1/s) / (1 + a^2/s^2) ]

= s / (s^2 + a^2)

and luckily this coincides with the expected Laplace transform of cos(at).

Source(s):

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