Skip to main content

Laplace transforms of functions

Short answer

For (cos2t - cos3t)/t, use ∫_s^∞ (u/(u^2+4) - u/(u^2+9)) du = ½ ln((s^2+9)/(s^2+4)). For sin^3 2t, rewrite using sin3θ identity: sin^3 2t = (3 sin2t - sin6t)/4, L = 48/(( s^2+4)(s^2+36)).

Find the Laplace transform of the following: a) (cos 2t − cos 3t)/t b) sin³ 2t.

a) L[(cos 2t − cos 3t)/t]

∫ₛ^∞ [u/(u² + 4) − u/(u² + 9)] du = ½ [ ln((u² + 4)/(u² + 9)) ]ₛ^∞
                                  = −½ ln((s² + 4)/(s² + 9))

Answer: ½ ln((s² + 9)/(s² + 4))

b) L[sin³ 2t]

Use sin 3θ = 3 sin θ − 4 sin³ θ, so sin³ θ = (3 sin θ − sin 3θ)/4. With θ = 2t:

sin³ 2t = (3 sin 2t − sin 6t)/4
L = ¼ [ 3·2/(s² + 4) − 6/(s² + 36) ]
  = (6/4) [ (s² + 36 − s² − 4) / ((s² + 4)(s² + 36)) ]
  = (3/2)·32 / ((s² + 4)(s² + 36))

Answer: 48/((s² + 4)(s² + 36))

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics III chapter 3 questions →