What is laplacetransform e^{-x}sin(x) ?

The answer to laplacetransform e^{-x}sin(x) is 1/((s+1)^2+1)

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laplacetransform e^{-x}sin(x)

laplace transform

e^{- x} sin (x)

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

Solution steps

L {e^{- x} sin (x)}

Apply transform rule:

L {f (t)} = F (s)

then

L {e^{a t} f (t)} = F (s - a)

For

e^{- x} sin (x) : f (x) = sin (x), a = - 1

= L {sin (x)} (s + 1)

L {sin (x)} : \frac{1}{s ^{2} + 1}

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

f (x) = \frac{e ^{x}}{5 - 2 x}

\frac{d}{d x} (e^{x s i n (2 x)})

x \to 0 lim (x^{7} e^{- x^{6}})

\frac{1 + s}{s ^{3} - 2 s ^{2}}

\frac{\partial}{\partial x} (21 y^{3} cos (7 y^{2} x))

What is laplacetransform e^{-x}sin(x) ?
The answer to laplacetransform e^{-x}sin(x) is 1/((s+1)^2+1)