What is laplacetransform sin(4t)cos(3t) ?

The answer to laplacetransform sin(4t)cos(3t) is 7/(2(s^2+49))+1/(2(s^2+1))

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laplacetransform sin(4t)cos(3t)

laplace transform

sin (4 t) cos (3 t)

\frac{7}{2 ( s ^{2} + 49 )} + \frac{1}{2 ( s ^{2} + 1 )}

Solution steps

L {sin (4 t) cos (3 t)}

Rewrite using trig identities

L {\frac{1}{2} sin (7 t) - \frac{1}{2} sin (- 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 {sin (7 t)} - \frac{1}{2} L {sin (- t)}

L {sin (7 t)} : \frac{7}{s ^{2} + 49}

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

= \frac{1}{2} \cdot \frac{7}{s ^{2} + 49} - \frac{1}{2} (- \frac{1}{s ^{2} + 1})

Refine

\frac{1}{2} \frac{7}{s ^{2} + 49} - \frac{1}{2} (- \frac{1}{s ^{2} + 1}) : \frac{7}{2 ( s ^{2} + 49 )} + \frac{1}{2 ( s ^{2} + 1 )}

= \frac{7}{2 ( s ^{2} + 49 )} + \frac{1}{2 ( s ^{2} + 1 )}

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

t + 3

\int 10 x - 3 d x

y = \frac{1}{x}, y = x, y = \frac{x}{4}

\frac{d y}{d x} = 1 - cos (2 x - y + 3)

What is laplacetransform sin(4t)cos(3t) ?
The answer to laplacetransform sin(4t)cos(3t) is 7/(2(s^2+49))+1/(2(s^2+1))