What is laplacetransform sin^2(t) ?

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

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laplacetransform sin^2(t)

laplace transform

sin^{2} (t)

\frac{2}{s ( s ^{2} + 4 )}

Solution steps

L {sin^{2} (t)}

Use the following identity:

sin^{2} (x) = \frac{1}{2} - cos (2 x) \frac{1}{2}

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

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 (2 t)}

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

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

= \frac{1}{2 s} - \frac{1}{2} \cdot \frac{s}{s ^{2} + 4}

Simplify

\frac{1}{2 s} - \frac{1}{2} \frac{s}{s ^{2} + 4} : \frac{2}{s ( s ^{2} + 4 )}

= \frac{2}{s ( s ^{2} + 4 )}

f (x) = 2 x^{3} - x^{2} + 9, (2, 21)

\int \frac{4}{35 sin ( x ) - 12 cos ( x )} d x

y^{^{''}} - 2 y^{^{'}} - 3 y = - 9 t e^{- t}

\int f x d x

\int (\frac{1}{5} - \frac{2}{x}) d x

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