What is laplacetransform e^{-t}cos^2(t) ?

The answer to laplacetransform e^{-t}cos^2(t) is 1/(2(s+1))+(s+1)/(2((s+1)^2+4))

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laplacetransform e^{-t}cos^2(t)

laplace transform

e^{- t} cos^{2} (t)

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

Solution steps

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

Rewrite using trig identities

L {\frac{1}{2} e^{- t} + \frac{1}{2} e^{- t} cos (2 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)}

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

L {e^{- t}} : \frac{1}{s + 1}

L {e^{- t} cos (2 t)} : \frac{s + 1}{( s + 1 ) ^{2} + 4}

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

Refine

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

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

\int \frac{x ^{2} + 8 x - 3}{x ^{3} + 3 x} d x

y^{^{''}} + 5 y^{^{'}} + 6 y = 108 x^{2}

\int_{7}^{\infty} \frac{ln ( x )}{x} d x

\int_{1}^{2} (x^{2}) d x

What is laplacetransform e^{-t}cos^2(t) ?
The answer to laplacetransform e^{-t}cos^2(t) is 1/(2(s+1))+(s+1)/(2((s+1)^2+4))