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Laplace transforms of functions coshatsinat

Short answer

Using cosh at = (e^(at)+e^(-at))/2 gives L[cosh at sin at] = a(s^2+2a^2)/(s^4+4a^4). Using integration property gives L[(cos 2t - cos 3t)/t] = (1/2) ln((s^2+9)/(s^2+4)).

Find the Laplace transform of: a) cosh at sin at b) (cos 2t − cos 3t)/t.

a) L[cosh at sin at]

cosh at = (e^(at) + e^(−at))/2 and L[sin at] = a/(s² + a²). By the first shift,

L[cosh at sin at] = ½ [ a/((s−a)² + a²) + a/((s+a)² + a²) ]

Denominators: (s − a)² + a² = s² − 2as + 2a² and (s + a)² + a² = s² + 2as + 2a²; the product is (s² + 2a²)² − 4a²s² = s⁴ + 4a⁴. The numerator is a[(s² + 2as + 2a²) + (s² − 2as + 2a²)] = 2a(s² + 2a²).

Answer: a(s² + 2a²)/(s⁴ + 4a⁴)

b) 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))

✓ Answer checked by Claude Opus 5.5 (high effort)

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