What is laplacetransform f(t)=cos^2(t) ?

The answer to laplacetransform f(t)=cos^2(t) is 1/(2s)+s/(2(s^2+4))

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laplacetransform f(t)=cos^2(t)

laplace transform

f (t) = cos^{2} (t)

\frac{1}{2 s} + \frac{s}{2 ( s ^{2} + 4 )}

Solution steps

L {cos^{2} (t)}

Use the following identity:

cos^{2} (x) = \frac{1}{2} + cos (2 x) \frac{1}{2}

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

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 (2 t)}

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

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

= \frac{1}{2 s} + \frac{1}{2} \cdot \frac{s}{s ^{2} + 4}

Refine

\frac{1}{2 s} + \frac{1}{2} \frac{s}{s ^{2} + 4} : \frac{1}{2 s} + \frac{s}{2 ( s ^{2} + 4 )}

= \frac{1}{2 s} + \frac{s}{2 ( s ^{2} + 4 )}

\frac{d}{d x} (3 x^{2} + 6)

\frac{d}{d x} (2 y sin (x) - x^{3} + ln (y))

\int_{- 1}^{1} e^{- x} d x

f (x) = \frac{1}{( 1 + x ^{2} )}, at x = - 1

\frac{5}{x}, at m

What is laplacetransform f(t)=cos^2(t) ?
The answer to laplacetransform f(t)=cos^2(t) is 1/(2s)+s/(2(s^2+4))