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

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

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laplacetransform e^{-t}cos(t)

laplace transform

e^{- t} cos (t)

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

Solution steps

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

Apply transform rule:

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

then

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

For

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

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

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

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

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

\int - \frac{x}{4} d x

\frac{d y}{d x}, x y^{2} = 5

\frac{d}{d x} (lo g_{4} (x^{3} - 7 x + 5))

\int_{0}^{π} - sin (x) d x

x \to 0 lim ((sin (x))^{t a n (x)})

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