What is laplacetransform-(sin^2(2t))/4 ?

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

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laplacetransform-(sin^2(2t))/4

laplace transform

- \frac{sin ^{2} ( 2 t )}{4}

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

Solution steps

L {- \frac{sin ^{2} ( 2 t )}{4}}

Rewrite using trig identities

L {- \frac{1}{8} + \frac{1}{8} 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}{8}} + \frac{1}{8} L {cos (4 t)}

L {\frac{1}{8}} : \frac{1}{8 s}

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

= - \frac{1}{8 s} + \frac{1}{8} \cdot \frac{s}{s ^{2} + 16}

Simplify

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

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

\frac{d}{d x} ((arctan (8 x))^{2})

\frac{d}{d x} (ln (1 + e^{x}))

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

x \to a lim (\frac{4}{1 - x})

f (x) = ln (x)

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