What is laplacetransform e^tcos(5t) ?

The answer to laplacetransform e^tcos(5t) is (s-1)/((s-1)^2+25)

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laplacetransform e^tcos(5t)

laplace transform

e^{t} cos (5 t)

\frac{s - 1}{( s - 1 ) ^{2} + 25}

Solution steps

L {e^{t} cos (5 t)}

Apply transform rule:

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

then

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

For

e^{t} cos (5 t) : f (t) = cos (5 t), a = 1

= L {cos (5 t)} (s - 1)

L {cos (5 t)} : \frac{s}{s ^{2} + 25}

= \frac{( s - 1 )}{( s - 1 ) ^{2} + 25}

= \frac{s - 1}{( s - 1 ) ^{2} + 25}

2 x (x + 1)^{3} (x^{2} + 2 x + 1)^{2}

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

\int \frac{3}{5 - 2 x} d x

f (x) = 3 sec (8 x)

\int cos^{6} (\frac{x}{2}) d x

What is laplacetransform e^tcos(5t) ?
The answer to laplacetransform e^tcos(5t) is (s-1)/((s-1)^2+25)