What is laplacetransform \H(t-2pi)sin(t-2pi) ?

The answer to laplacetransform \H(t-2pi)sin(t-2pi) is (e^{-2pis})/(s^2+1)

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laplacetransform \H(t-2pi)sin(t-2pi)

laplace transform

H (t - 2 π) sin (t - 2 π)

\frac{e ^{- 2 π s}}{s ^{2} + 1}

Solution steps

L {H (t - 2 π) sin (t - 2 π)}

Use Laplace Transform table:

L {H (t - c) f (t)} = e^{- cs} L {f (t + c)}

For

H (t - 2 π) sin (t - 2 π) : f (t) = sin (t - 2 π), c = 2 π

= e^{- 2 π s} L {sin ((t + 2 π) - 2 π)}

L {sin ((t + 2 π) - 2 π)} : \frac{1}{s ^{2} + 1}

= e^{- 2 π s} \frac{1}{s ^{2} + 1}

= \frac{e ^{- 2 π s}}{s ^{2} + 1}

\frac{d}{d x} (\frac{2 + x}{( 2 - x ) ^{5}})

f (x) = 73 sin (x) + 7 cos (x)

\int_{1}^{5} \frac{M}{e ^{M}} d M

\int \frac{1}{x ( a + b x )} d x

\frac{d y}{d t} + e^{t} y = - 20 e^{t}

What is laplacetransform \H(t-2pi)sin(t-2pi) ?
The answer to laplacetransform \H(t-2pi)sin(t-2pi) is (e^{-2pis})/(s^2+1)