What is laplacetransform 8sin(t) ?

The answer to laplacetransform 8sin(t) is 8/(s^2+1)

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laplacetransform 8sin(t)

laplace transform

8 sin (t)

\frac{8}{s ^{2} + 1}

Solution steps

L {8 sin (t)}

Use the constant multiplication property of Laplace Transform:

For function

f (t)

and constant

a : L {a \cdot f (t)} = a \cdot L {f (t)}

= 8 L {sin (t)}

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

= 8 \cdot \frac{1}{s ^{2} + 1}

Refine

8 \frac{1}{s ^{2} + 1} : \frac{8}{s ^{2} + 1}

= \frac{8}{s ^{2} + 1}

k = 1 \sum \infty \frac{k ^{3}}{5 ^{k}}

\int \frac{7 sin ( x ) + 2 sec ( x )}{tan ( x )} d x

f^{^{'}} (x) = 22 x + ln (22 x)

\frac{s}{2 s + 4}

\frac{\partial}{\partial x} (\frac{5}{x y} + \frac{x}{y} + \frac{y}{x})

What is laplacetransform 8sin(t) ?
The answer to laplacetransform 8sin(t) is 8/(s^2+1)