What is laplacetransform (sin(t))^2 ?

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

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laplacetransform (sin(t))^2

laplace transform

(sin (t))^{2}

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

Solution steps

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

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 )}

\int 2 t \cdot (- cos (2 t)) d t

f (x) = e^{x}, at x = 5

x \to 0 lim (\frac{2}{1 - 2 ^{\frac{1}{x}}})

t^{3} \frac{d y}{d t} + 4 t^{2} y = e^{- t}, y (- 1) = 0

f (x) = \frac{2}{x - 1}

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