What is laplacetransform t(tcos(t)-sin(t)) ?

The answer to laplacetransform t(tcos(t)-sin(t)) is -(8s)/((s^2+1)^3)

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laplacetransform t(tcos(t)-sin(t))

laplace transform

t (t cos (t) - sin (t))

- \frac{8 s}{( s ^{2} + 1 ) ^{3}}

Solution steps

L {t (t cos (t) - sin (t))}

Use Laplace Transform table:

L {t^{k} f (t)} = (- 1)^{k} \frac{d ^{k}}{d s ^{k}} (L {f (t)})

For

t (t cos (t) - sin (t)) : f (t) = (t cos (t) - sin (t)), k = 1

= (- 1)^{1} \frac{d}{d s} (L {(t cos (t) - sin (t))})

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

\frac{d}{d s} (- \frac{2}{( s ^{2} + 1 ) ^{2}}) = \frac{8 s}{( s ^{2} + 1 ) ^{3}}

= (- 1)^{1} \frac{8 s}{( s ^{2} + 1 ) ^{3}}

= - \frac{8 s}{( s ^{2} + 1 ) ^{3}}

n = 1 \sum \infty \frac{- 1 ^{n}}{n}

x \to 2 lim (\frac{x ^{2} - 3 x + 2}{x ^{2} + 5 x - 14})

t^{2} \cdot e^{- 4 t}

\int sin^{4} (x) cos^{2} (x) d x

(22, 54), (62, 27)

What is laplacetransform t(tcos(t)-sin(t)) ?
The answer to laplacetransform t(tcos(t)-sin(t)) is -(8s)/((s^2+1)^3)