What is laplacetransform cos^2(2t) ?

The answer to laplacetransform cos^2(2t) is 1/(2s)+s/(2(s^2+16))

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laplacetransform cos^2(2t)

laplace transform

cos^{2} (2 t)

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

Solution steps

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

Rewrite using trig identities

L {\frac{1}{2} + \frac{1}{2} 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}{2}} + \frac{1}{2} L {cos (4 t)}

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

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

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

Refine

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

= \frac{1}{2 s} + \frac{s}{2 ( s ^{2} + 16 )}

\int_{- 2}^{2} (12 - 3 x^{2}) d x

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\frac{d}{d x} ((\frac{d}{d x} (ln (cos (x))))^{2})

\int \frac{x - 39}{x ^{2} - x - 12} d x

\int_{0}^{1} \frac{4 t ^{3}}{1 + t ^{4}} d t

What is laplacetransform cos^2(2t) ?
The answer to laplacetransform cos^2(2t) is 1/(2s)+s/(2(s^2+16))