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) \cdot 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 x sin^{2} (4 x) d x

y = ∣ x - 3 ∣, y = \frac{x}{2}, y = 0

\int 7 x^{2} e^{x} d x

4 x^{5} ln (x)

\frac{d}{d x} (3 - 4 x)

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