sinh at cos bt is transformed by

writing sinh at = (e^(at) - e^(-at))/2 and using the first shift

(e^(-at) - e^(-bt))/t uses

integral of F(u) from s to infinity

Short answer

(i) sinh(at)cos(bt) transforms to a(s^2 - a^2 - b^2) / [((s-a)^2+b^2)((s+a)^2+b^2)]. (ii) (e^(-at)-e^(-bt))/t transforms to ln((s+b)/(s+a)).

Find the Laplace transform of: (i) sinh at cos bt (ii) (e^(−at) − e^(−bt))/t.

sinh at = (e^(at) − e^(−at))/2 and L[cos bt] = s/(s² + b²). First shifting theorem: L[e^(at) cos bt] = (s − a)/((s − a)² + b²).

L[sinh at cos bt] = ½ [ (s − a)/((s − a)² + b²) − (s + a)/((s + a)² + b²) ]

The numerator after combining: (s − a)((s + a)² + b²) − (s + a)((s − a)² + b²) = 2a(s² − a² − b²).

Answer: a(s² − a² − b²) / [ ((s − a)² + b²)((s + a)² + b²) ]

∫ₛ^∞ [1/(u + a) − 1/(u + b)] du = [ ln((u + a)/(u + b)) ]ₛ^∞ = −ln((s + a)/(s + b))

Answer: ln((s + b)/(s + a))

✓ Answer checked by Claude Opus 5.5 (high effort)