What is laplacetransform sin(1/2 t)*t^2 ?

The answer to laplacetransform sin(1/2 t)*t^2 is -(16(1-12s^2))/((4s^2+1)^3)

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laplacetransform sin(1/2 t)*t^2

laplace transform

sin (\frac{1}{2} t) \cdot t^{2}

- \frac{16 ( 1 - 12 s ^{2} )}{( 4 s ^{2} + 1 ) ^{3}}

Solution steps

L {t^{2} sin (\frac{1}{2} t)}

Use Laplace Transform table:

L {t^{k} f (t)} = (- 1)^{k} \frac{d ^{k}}{d s ^{k}} (L {f (t)})

For

sin (\frac{1}{2} t) t^{2} : f (t) = sin (\frac{1}{2} t), k = 2

= (- 1)^{2} \frac{d ^{2}}{d s ^{2}} (L {sin (\frac{1}{2} t)})

L {sin (\frac{1}{2} t)} : \frac{2}{4 s ^{2} + 1}

\frac{d ^{2}}{d s ^{2}} (\frac{2}{4 s ^{2} + 1}) = - \frac{16 ( 1 - 12 s ^{2} )}{( 4 s ^{2} + 1 ) ^{3}}

= (- 1)^{2} (- \frac{16 ( 1 - 12 s ^{2} )}{( 4 s ^{2} + 1 ) ^{3}})

= - \frac{16 ( 1 - 12 s ^{2} )}{( 4 s ^{2} + 1 ) ^{3}}

What is laplacetransform sin(1/2 t)*t^2 ?
The answer to laplacetransform sin(1/2 t)*t^2 is -(16(1-12s^2))/((4s^2+1)^3)