Solve L(sin4t) ?

Question

5702 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-12T22:59:23

Guessing this is a Laplace transform request

$mathbb L {cos omega t color(red)(-) i sin omega t } = mathbb L {e^(-i omega t) }$

$= int_0^(oo) e^(-(i omega +s)t) \ dt$

$= - (1)/(i omega + s) [ e^(-(i omega +s)t)]_0^(oo)$

$= (1)/(i omega + s) = (s - i omega)/(s^2 + omega ^2)$

$implies mathbb L {cos omega t } = \mathcal (Re) (mathbb L {cos omega t color(red)(-) i sin omega t } )= (s )/(s^2 + omega ^2)$

AND

$implies mathbb L {sin omega t } = -\mathcal (Im) (mathbb L {cos omega t color(red)(-) i sin omega t } )= (omega )/(s^2 + omega ^2)$

Repeat for: $omega = 4$