What is laplacetransform 4cos^2(2t) ?

The answer to laplacetransform 4cos^2(2t) is 2/s+(2s)/(s^2+16)

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laplacetransform 4cos^2(2t)

laplace transform

4 cos^{2} (2 t)

\frac{2}{s} + \frac{2 s}{s ^{2} + 16}

Solution steps

L {4 cos^{2} (2 t)}

Rewrite using trig identities

L {2 + 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 {2} + 2 L {cos (4 t)}

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

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

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

Refine

\frac{2}{s} + 2 \frac{s}{s ^{2} + 16} : \frac{2}{s} + \frac{2 s}{s ^{2} + 16}

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

\frac{3 x + 7}{4 x - 5}

n = 0 \sum \infty 37.06

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

\frac{d}{d x} (\frac{x ^{3} - 8}{x ^{2} - 4})

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

What is laplacetransform 4cos^2(2t) ?
The answer to laplacetransform 4cos^2(2t) is 2/s+(2s)/(s^2+16)