What is laplacetransform cos^2(16t) ?

The answer to laplacetransform cos^2(16t) is 1/(2s)+s/(2(s^2+1024))

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laplacetransform cos^2(16t)

laplace transform

cos^{2} (16 t)

\frac{1}{2 s} + \frac{s}{2 ( s ^{2} + 1024 )}

Solution steps

L {cos^{2} (16 t)}

Rewrite using trig identities

L {\frac{1}{2} + \frac{1}{2} cos (32 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 (32 t)}

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

L {cos (32 t)} : \frac{s}{s ^{2} + 1024}

= \frac{1}{2 s} + \frac{1}{2} \cdot \frac{s}{s ^{2} + 1024}

Refine

\frac{1}{2 s} + \frac{1}{2} \frac{s}{s ^{2} + 1024} : \frac{1}{2 s} + \frac{s}{2 ( s ^{2} + 1024 )}

= \frac{1}{2 s} + \frac{s}{2 ( s ^{2} + 1024 )}

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

\frac{d x}{d t} = e^{x} \cdot sin (t)

sin (\frac{x π}{4})

(x^{- \frac{1}{2}})^{^{'}}

x \to \infty lim (\frac{1}{0})

What is laplacetransform cos^2(16t) ?
The answer to laplacetransform cos^2(16t) is 1/(2s)+s/(2(s^2+1024))