What is laplacetransform sin^2(3t) ?

The answer to laplacetransform sin^2(3t) is (18)/(s(s^2+36))

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laplacetransform sin^2(3t)

laplace transform

sin^{2} (3 t)

\frac{18}{s ( s ^{2} + 36 )}

Solution steps

L {sin^{2} (3 t)}

Rewrite using trig identities

L {\frac{1}{2} - \frac{1}{2} cos (6 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 (6 t)}

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

L {cos (6 t)} : \frac{s}{s ^{2} + 36}

= \frac{1}{2 s} - \frac{1}{2} \cdot \frac{s}{s ^{2} + 36}

Simplify

\frac{1}{2 s} - \frac{1}{2} \frac{s}{s ^{2} + 36} : \frac{18}{s ( s ^{2} + 36 )}

= \frac{18}{s ( s ^{2} + 36 )}

\frac{\partial}{\partial x} (2 x^{3} + 6 x y - 2 y^{4})

\frac{x ^{2} + 1}{2 x - 3}

x^{2} = \frac{4 x + y}{4 x - y}

\int \frac{2 x}{4 - x ^{4}} d x

\int \frac{x + 4}{x ^{3}} d x

What is laplacetransform sin^2(3t) ?
The answer to laplacetransform sin^2(3t) is (18)/(s(s^2+36))