Problem 02
Find the Laplace transform of $f(t) = \sin^2 t$ using the transform of derivatives.
Solution 02
$f(t) = \sin^2 t$ .......... $f(0) = 0$
$f'(t) = 2\sin t \, \cos t = \sin 2t$
$\mathcal{L} \left\{ f'(t) \right\} = s \, \mathcal{L} \left\{ f(t) \right\} - f(0)$
$\mathcal{L} (\sin 2t) = s \, \mathcal{L} (\sin^2 t) - 0$
$\dfrac{2}{s^2 + 2^2} = s \, \mathcal{L} (\sin^2 t)$
$s \, \mathcal{L} (\sin^2 t) = \dfrac{2}{s^2 + 4}$