What is laplacetransform (cos(4t)-8)^2 ?

The answer to laplacetransform (cos(4t)-8)^2 is 1/(2s)+s/(2(s^2+64))-(16s}{s^2+16}+\frac{64)/s

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laplacetransform (cos(4t)-8)^2

laplace transform

(cos (4 t) - 8)^{2}

\frac{1}{2 s} + \frac{s}{2 ( s ^{2} + 64 )} - \frac{16 s}{s ^{2} + 16} + \frac{64}{s}

Solution steps

L {(cos (4 t) - 8)^{2}}

Expand

(cos (4 t) - 8)^{2} : cos^{2} (4 t) - 16 cos (4 t) + 64

= L {cos^{2} (4 t) - 16 cos (4 t) + 64}

Rewrite using trig identities

L {\frac{1}{2} + \frac{1}{2} cos (8 t) - 16 cos (4 t) + 64}

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 (8 t)} - 16 L {cos (4 t)} + L {64}

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

L {cos (8 t)} : \frac{s}{s ^{2} + 64}

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

L {64} : \frac{64}{s}

= \frac{1}{2 s} + \frac{1}{2} \cdot \frac{s}{s ^{2} + 64} - 16 \cdot \frac{s}{s ^{2} + 16} + \frac{64}{s}

Refine

\frac{1}{2 s} + \frac{1}{2} \frac{s}{s ^{2} + 64} - 16 \frac{s}{s ^{2} + 16} + \frac{64}{s} : \frac{1}{2 s} + \frac{s}{2 ( s ^{2} + 64 )} - \frac{16 s}{s ^{2} + 16} + \frac{64}{s}

= \frac{1}{2 s} + \frac{s}{2 ( s ^{2} + 64 )} - \frac{16 s}{s ^{2} + 16} + \frac{64}{s}

\frac{d y}{d x} + 6 tan (6 x) y = - 8 cos (6 x)

\int \frac{x - 1}{x ^{2} + 2 x} d x

(1 - x^{2})

\int 8 \frac{tan ^{3} ( 2 x )}{x} d x

n = 2 \sum \infty n (n - 1) x^{n}

What is laplacetransform (cos(4t)-8)^2 ?
The answer to laplacetransform (cos(4t)-8)^2 is 1/(2s)+s/(2(s^2+64))-(16s}{s^2+16}+\frac{64)/s