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

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

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

laplace transform

sin (t) H (t - π)

- \frac{e ^{- π s}}{s ^{2} + 1}

Solution steps

L {sin (t) H (t - π)}

Use Laplace Transform table:

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

For

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

= e^{- π s} L {sin ((t + π))}

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

= e^{- π s} (- \frac{1}{s ^{2} + 1})

= - \frac{e ^{- π s}}{s ^{2} + 1}

x^{3}, 0, 2

\frac{\partial}{\partial s} (- s^{2} t^{2})

\frac{d}{d x} (cot (x) sec (x))

\frac{d}{d t} (s cos (t))

\frac{d y}{d x}, x^{y} = y^{x}

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