What is laplacetransform sin^2(4t) ?

The answer to laplacetransform sin^2(4t) is (32)/(s(s^2+64))

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laplacetransform sin^2(4t)

laplace transform

sin^{2} (4 t)

\frac{32}{s ( s ^{2} + 64 )}

Solution steps

L {sin^{2} (4 t)}

Rewrite using trig identities

L {\frac{1}{2} - \frac{1}{2} cos (8 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 (8 t)}

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

L {cos (8 t)} : \frac{s}{s ^{2} + 64}

= \frac{1}{2 s} - \frac{1}{2} \cdot \frac{s}{s ^{2} + 64}

Simplify

\frac{1}{2 s} - \frac{1}{2} \frac{s}{s ^{2} + 64} : \frac{32}{s ( s ^{2} + 64 )}

= \frac{32}{s ( s ^{2} + 64 )}

\frac{d}{d x} (4 x^{2} - 3 x + 1)

\frac{\partial}{\partial x} (2 \frac{y}{x})

x \to 0 lim (e^{4 x})

\frac{d x}{d t} = 6.4286 - 0.10714 x

y = 5^{x^{3}}

What is laplacetransform sin^2(4t) ?
The answer to laplacetransform sin^2(4t) is (32)/(s(s^2+64))