Problem 02 | Laplace Transform of Derivatives

$f(t) = \sin^2 t$       ..........       $f(0) = 0$

$f'(t) = 2\sin t \, \cos t = \sin 2t$
 

$\mathcal{L} \left\{ f'(t) \right\} = s \, \mathcal{L} \left\{ f(t) \right\} - f(0)$

$\mathcal{L} (\sin 2t) = s \, \mathcal{L} (\sin^2 t) - 0$

$\dfrac{2}{s^2 + 2^2} = s \, \mathcal{L} (\sin^2 t)$

$s \, \mathcal{L} (\sin^2 t) = \dfrac{2}{s^2 + 4}$

$\mathcal{L} (\sin^2 t) = \dfrac{2}{s(s^2 + 4)}$           answer