Find the Laplace transform of the following functions:(i) sin 2t cos 3t

The Laplace transform of a function f(t) is defined as:

L{f(t)} = ∫_0^∞ e^(-st)f(t) dt

To find the Laplace transform of the function sin(2t)cos(3t), we can use the product-to-sum identities from trigonometry to simplify the function. The identity is:

sin(a)cos(b) = 1/2[sin(a-b) + sin(a+b)]

Applying this identity to sin(2t)cos(3t) gives:

sin(2t)cos(3t) = 1/2[sin(2t-3t) + sin(2t+3t)] = 1/2[sin(-t) + sin(5t)]

Now, we can find the Laplace transform of each term separately. The Laplace transform of sin(at) is a/(s^2 + a^2). So, we have:

L{sin(-t)} = -1/(s^2 + 1) L{sin(5t)} = 5/(s^2 + 25)

Therefore, the Laplace transform of the original function is:

L{sin(2t)cos(3t)} = 1/2[L{sin(-t)} + L{sin(5t)}] = 1/2[-1/(s^2 + 1) + 5/(s^2 + 25)]