What is laplacetransform cos^2(6t) ?

The answer to laplacetransform cos^2(6t) is 1/(2s)+s/(2(s^2+144))

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laplacetransform cos^2(6t)

laplace transform

cos^{2} (6 t)

\frac{1}{2 s} + \frac{s}{2 ( s ^{2} + 144 )}

Solution steps

L {cos^{2} (6 t)}

Rewrite using trig identities

L {\frac{1}{2} + \frac{1}{2} cos (12 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 (12 t)}

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

L {cos (12 t)} : \frac{s}{s ^{2} + 144}

= \frac{1}{2 s} + \frac{1}{2} \cdot \frac{s}{s ^{2} + 144}

Refine

\frac{1}{2 s} + \frac{1}{2} \frac{s}{s ^{2} + 144} : \frac{1}{2 s} + \frac{s}{2 ( s ^{2} + 144 )}

= \frac{1}{2 s} + \frac{s}{2 ( s ^{2} + 144 )}

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

f (x) = lo g_{\frac{3}{2}} (x) + 2

\int \frac{( ln ( x ) ) ^{38}}{x} d x

\int 1 - t^{2} d t

g (x) = \frac{x ^{2} - 16}{x - 4}

What is laplacetransform cos^2(6t) ?
The answer to laplacetransform cos^2(6t) is 1/(2s)+s/(2(s^2+144))