What is laplacetransform e^{-0.4t}cos(12t) ?

The answer to laplacetransform e^{-0.4t}cos(12t) is (s+0.4)/((s+0.4)^2+144)

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laplacetransform e^{-0.4t}cos(12t)

laplace transform

e^{- 0.4 t} cos (12 t)

\frac{s + 0.4}{( s + 0.4 ) ^{2} + 144}

Solution steps

L {e^{- 0.4 t} cos (12 t)}

Apply transform rule:

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

then

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

For

e^{- 0.4 t} cos (12 t) : f (t) = cos (12 t), a = - 0.4

= L {cos (12 t)} (s + 0.4)

L {cos (12 t)} : \frac{s}{s ^{2} + 144}

= \frac{( s + 0.4 )}{( s + 0.4 ) ^{2} + 144}

= \frac{s + 0.4}{( s + 0.4 ) ^{2} + 144}

\int \frac{2 x}{x ^{2} + 4} d x

T \to \infty lim (f (T))

\int sec^{5} (3 x) tan (3 x) d x

\frac{x}{y}

\int \frac{1}{( x ^{2} - 144 ) ^{\frac{3}{2}}} d x

What is laplacetransform e^{-0.4t}cos(12t) ?
The answer to laplacetransform e^{-0.4t}cos(12t) is (s+0.4)/((s+0.4)^2+144)