Answer: To find the Laplace transform of e^(1/2) sin(t) * cos(4t), we use the properties of the Laplace transform and the formula for the Laplace transform of sin(at) and cos(bt).
Using the formula for the Laplace transform of sin(at):
L{sin(at)} = a / (s^2 + a^2)
And the formula for the Laplace transform of cos(bt):
L{cos(bt)} = s / (s^2 + b^2)
We can write e^(1/2) sin(t) * cos(4t) as:
e^(1/2) * sin(t) * cos(4t) = (1/2) * (e^(1/2) * sin(5t) - e^(1/2) * sin(3t))
To find the Laplace transform of each term, we can use linearity property of the Laplace transform:
L{e^(1/2) * sin(t) * cos(4t)} = (1/2) * (L{e^(1/2) * sin(5t)} - L{e^(1/2) * sin(3t)})
Applying the Laplace transform to each term:
L{e^(1/2) * sin(5t)} = 1 / (s^2 + (5/2)^2)
L{e^(1/2) * sin(3t)} = 1 / (s^2 + (3/2)^2)
So the Laplace transform of e^(1/2) sin(t) * cos(4t) is:
L{e^(1/2) * sin(t) * cos(4t)} = (1/2) * (1 / (s^2 + (5/2)^2) - 1 / (s^2 + (3/2)^2))