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Laplace Transform of sin^2(t)
Published on October 27, 2024
Calculus
The Laplace transform of sin^2(t) is 2 / (s * (s^2 + 4)).
Question
laplace transform sin^2(t)
Answer
Let’s find the Laplace transform of sin^2(t).
1. Utilize Trigonometric Identities
Recall the double-angle identity for cosine:
cos(2t) = 1 - 2sin^2(t)
cos(2t) = 1 - 2sin^2(t)
Solving for sin^2(t), we get:
sin^2(t) = (1 - cos(2t))/2
sin^2(t) = (1 - cos(2t))/2
2. Apply Linearity
The Laplace transform is linear. This means:
L[af(t) + bg(t)] = aL[f(t)] + bL[g(t)]
L[af(t) + bg(t)] = aL[f(t)] + bL[g(t)]
Applying this to our problem:
L[sin^2(t)] = L[(1 - cos(2t))/2] = (1/2)L[1] - (1/2)L[cos(2t)]
L[sin^2(t)] = L[(1 - cos(2t))/2] = (1/2)L[1] - (1/2)L[cos(2t)]
3. Use Laplace Transform Table
Now we can use the standard Laplace transform table:
- L[1] = 1/s
- L[cos(at)] = s/(s^2 + a^2)
4. Substitute and Simplify
Substituting these into our expression:
(1/2)L[1] - (1/2)L[cos(2t)] = (1/2)(1/s) - (1/2)(s/(s^2 + 2^2))
Simplifying:
= 1/(2s) - s/(2(s^2 + 4))
= (s^2 + 4 - s^2) / (2s(s^2 + 4))
= 2/(s(s^2 + 4))
Therefore, the Laplace transform of sin^2(t) is 2/(s(s^2 + 4)).