What is inverse oflaplace s/(s^2+1) ?

The answer to inverse oflaplace s/(s^2+1) is cos(t)

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inverse oflaplace s/(s^2+1)

inverse laplace

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

cos (t)

Solution steps

L^{- 1} {\frac{s}{s ^{2} + 1}}

Apply inverse transform rule:

L^{- 1} {F (s)} = f (t)

then

L^{- 1} {s F (s)} = \frac{d}{d t} f (t) + f (0)

For

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

= \frac{d}{d t} (L^{- 1} {\frac{1}{s ^{2} + 1}}) + L^{- 1} {\frac{1}{s ^{2} + 1}} (0)

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

\frac{d}{d t} (sin (t)) = cos (t)

Evaluate

sin (t)

at point

0 = 0

= cos (t)

\int \frac{1}{36 - 6 t ^{2}} d t

f (x) = 2 - x^{2}

{\frac{1}{S ^{3}}}

\int I n (x) d x

ln (1 - 2 x)

What is inverse oflaplace s/(s^2+1) ?
The answer to inverse oflaplace s/(s^2+1) is cos(t)