What is laplacetransform e^{-3t}cos(-t) ?

The answer to laplacetransform e^{-3t}cos(-t) is (s+3)/((s+3)^2+1)

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laplacetransform e^{-3t}cos(-t)

laplace transform

e^{- 3 t} cos (- t)

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

Solution steps

L {e^{- 3 t} cos (- t)}

Apply transform rule:

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

then

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

For

e^{- 3 t} cos (- t) : f (t) = cos (- t), a = - 3

= L {cos (- t)} (s + 3)

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

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

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

\int 5 u^{\frac{3}{2}} d u

\frac{d}{d x} (\frac{2 x - 6}{4 - x})

\int \frac{x ^{2}}{( x ^{3} + 1 ) ^{2}} d x

\frac{d}{d x} (\frac{9}{x + 4})

\frac{d}{d x} (\frac{2 x + 5}{7 x ^{2}})

What is laplacetransform e^{-3t}cos(-t) ?
The answer to laplacetransform e^{-3t}cos(-t) is (s+3)/((s+3)^2+1)