What is laplacetransform cos(4t)cos(6t) ?

The answer to laplacetransform cos(4t)cos(6t) is s/(2(s^2+100))+s/(2(s^2+4))

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laplacetransform cos(4t)cos(6t)

laplace transform

cos (4 t) cos (6 t)

\frac{s}{2 ( s ^{2} + 100 )} + \frac{s}{2 ( s ^{2} + 4 )}

Solution steps

L {cos (4 t) cos (6 t)}

Rewrite using trig identities

L {\frac{1}{2} cos (10 t) + \frac{1}{2} cos (- 2 t)}

Use the linearity property of Laplace Transform:

For functions

f (t), g (t)

and constants

a, b : L {a \cdot f (t) + b \cdot g (t)} = a \cdot L {f (t)} + b \cdot L {g (t)}

= \frac{1}{2} L {cos (10 t)} + \frac{1}{2} L {cos (- 2 t)}

L {cos (10 t)} : \frac{s}{s ^{2} + 100}

L {cos (- 2 t)} : \frac{s}{s ^{2} + 4}

= \frac{1}{2} \cdot \frac{s}{s ^{2} + 100} + \frac{1}{2} \cdot \frac{s}{s ^{2} + 4}

Refine

\frac{1}{2} \frac{s}{s ^{2} + 100} + \frac{1}{2} \frac{s}{s ^{2} + 4} : \frac{s}{2 ( s ^{2} + 100 )} + \frac{s}{2 ( s ^{2} + 4 )}

= \frac{s}{2 ( s ^{2} + 100 )} + \frac{s}{2 ( s ^{2} + 4 )}

\int 5 sin^{3} (x) cos (x) d x

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

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

\int \frac{1 - x}{x ^{2} + 1} d x

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

What is laplacetransform cos(4t)cos(6t) ?
The answer to laplacetransform cos(4t)cos(6t) is s/(2(s^2+100))+s/(2(s^2+4))