Question: Find the Laplace transform for f(x)=(sin3x-cos2x).
Simply use the table of Laplace transforms.
L{sin(at)} = a/(s^2 + a^2)
L{cos(at)} = s/(s2 + a^2)
So, L{sin(3x)} = 3/(s^2 + 9)
And L{cos(2x)} = s/(s^2 + 4)
Then, L{f(x)} = 3/(s^2 + 9) - s/(s^2 + 4) = (3s^2 + 12 - s^3 - 9s)/{(s^2 + 9)(s^2 + 4)}
Answer: L{f(x)} = -(s^3 - 3s^2 + 9s - 12)/{(s^2 + 9)(s^2 + 4)}