What is laplacetransform h(t-2pi)sin(t-2pi) ?

The answer to laplacetransform h(t-2pi)sin(t-2pi) is (2hs)/((s^2+1)^2)-(2pih)/(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{2 h s}{( s ^{2} + 1 ) ^{2}} - \frac{2 πh}{s ^{2} + 1}

Solution steps

L {h sin (t - 2 π) (t - 2 π)}

Expand

h (t - 2 π) sin (t - 2 π) : h t sin (t - 2 π) - 2 πh sin (t - 2 π)

= L {h t sin (t - 2 π) - 2 πh sin (t - 2 π)}

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)}

= h L {t sin (t - 2 π)} - 2 πh L {sin (t - 2 π)}

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

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

= h \frac{2 s}{( s ^{2} + 1 ) ^{2}} - 2 πh \frac{1}{s ^{2} + 1}

Refine

h \frac{2 s}{( s ^{2} + 1 ) ^{2}} - 2 πh \frac{1}{s ^{2} + 1} : \frac{2 h s}{( s ^{2} + 1 ) ^{2}} - \frac{2 πh}{s ^{2} + 1}

= \frac{2 h s}{( s ^{2} + 1 ) ^{2}} - \frac{2 πh}{s ^{2} + 1}

\frac{d}{d x} (x^{4} y)

\frac{d}{d x} (\frac{1}{x ^{2} y})

\int_{0}^{1} x^{2} (x + 1)^{7} d x

\frac{\partial}{\partial y} (2 x y^{3} - 4)

y^{^{''}} - 18 y^{^{'}} + 81 y = 0

What is laplacetransform h(t-2pi)sin(t-2pi) ?
The answer to laplacetransform h(t-2pi)sin(t-2pi) is (2hs)/((s^2+1)^2)-(2pih)/(s^2+1)