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

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

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laplacetransform e^{-t}sin(t)

laplace transform

e^{- t} sin (t)

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

Solution steps

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

Apply transform rule:

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

then

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

For

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

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

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

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

x \to \infty lim (lo g_{e} (x^{2} - 1))

x \to 3 lim (f (x))

x y^{^{'}} + y = cos (x)

\int \frac{4 x \frac{5}{4} - 2 x \frac{1}{4}}{x \frac{1}{4}} d x

\int_{0}^{\frac{π}{2}} (3 cos^{2} (x)) d x

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