What is laplacetransform sin(2t)cos(5t) ?

The answer to laplacetransform sin(2t)cos(5t) is 7/(2(s^2+49))-3/(2(s^2+9))

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laplacetransform sin(2t)cos(5t)

laplace transform

sin (2 t) cos (5 t)

\frac{7}{2 ( s ^{2} + 49 )} - \frac{3}{2 ( s ^{2} + 9 )}

Solution steps

L {sin (2 t) cos (5 t)}

Rewrite using trig identities

L {\frac{1}{2} sin (7 t) - \frac{1}{2} sin (3 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 (3 t)}

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

L {sin (3 t)} : \frac{3}{s ^{2} + 9}

= \frac{1}{2} \cdot \frac{7}{s ^{2} + 49} - \frac{1}{2} \cdot \frac{3}{s ^{2} + 9}

Refine

\frac{1}{2} \frac{7}{s ^{2} + 49} - \frac{1}{2} \frac{3}{s ^{2} + 9} : \frac{7}{2 ( s ^{2} + 49 )} - \frac{3}{2 ( s ^{2} + 9 )}

= \frac{7}{2 ( s ^{2} + 49 )} - \frac{3}{2 ( s ^{2} + 9 )}

\int \frac{- 3 x ^{2}}{1 - x} d x

\int 8 ln (x^{2} - 1) d x

\frac{\partial}{\partial y} (z arctan (\frac{y}{x}))

f (x) = \frac{1}{x ln ( x )}

\int_{0}^{1} 64 t^{2} + 9 t^{4} d t

What is laplacetransform sin(2t)cos(5t) ?
The answer to laplacetransform sin(2t)cos(5t) is 7/(2(s^2+49))-3/(2(s^2+9))