What is laplacetransform sin^2(2t) ?

The answer to laplacetransform sin^2(2t) is 8/(s(s^2+16))

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laplacetransform sin^2(2t)

laplace transform

sin^{2} (2 t)

\frac{8}{s ( s ^{2} + 16 )}

Solution steps

L {sin^{2} (2 t)}

Rewrite using trig identities

L {\frac{1}{2} - \frac{1}{2} cos (4 t)}

Use the linearity property of Laplace Transform:

For functions

f (t), g (t)

and constants

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= L {\frac{1}{2}} - \frac{1}{2} L {cos (4 t)}

L {\frac{1}{2}} : \frac{1}{2 s}

L {cos (4 t)} : \frac{s}{s ^{2} + 16}

= \frac{1}{2 s} - \frac{1}{2} \cdot \frac{s}{s ^{2} + 16}

Simplify

\frac{1}{2 s} - \frac{1}{2} \frac{s}{s ^{2} + 16} : \frac{8}{s ( s ^{2} + 16 )}

= \frac{8}{s ( s ^{2} + 16 )}

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What is laplacetransform sin^2(2t) ?
The answer to laplacetransform sin^2(2t) is 8/(s(s^2+16))