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

The answer to laplacetransform sin(t)sin(3t) is (6s)/((s^2+4)(s^2+16))

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laplacetransform sin(t)sin(3t)

laplace transform

sin (t) sin (3 t)

\frac{6 s}{( s ^{2} + 4 ) ( s ^{2} + 16 )}

Solution steps

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

Rewrite using trig identities

L {- \frac{1}{2} cos (4 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 (4 t)} + \frac{1}{2} L {cos (- 2 t)}

L {cos (4 t)} : \frac{s}{s ^{2} + 16}

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

= - \frac{1}{2} \cdot \frac{s}{s ^{2} + 16} + \frac{1}{2} \cdot \frac{s}{s ^{2} + 4}

Simplify

- \frac{1}{2} \frac{s}{s ^{2} + 16} + \frac{1}{2} \frac{s}{s ^{2} + 4} : \frac{6 s}{( s ^{2} + 4 ) ( s ^{2} + 16 )}

= \frac{6 s}{( s ^{2} + 4 ) ( s ^{2} + 16 )}

\int cos^{2} (\frac{x}{4}) d x

\int \frac{1}{x + 9} d x

\int \frac{4}{x ^{2} + 5 x - 14} d x

\int ln (14 x) d x

x \to 1 lim (cos (π x))

What is laplacetransform sin(t)sin(3t) ?
The answer to laplacetransform sin(t)sin(3t) is (6s)/((s^2+4)(s^2+16))