Question

find laplace transform of attached question ​

Answer 1

Answer:

Use integral definition of laplace transform to get

∫∞0sinh(at)exp(−st)dt

=limb→∞∫b0sinh(at)exp(−st)dt

By the definition sinh(at)=12(exp(at)−exp(−at))

We can write

=limb→∞12∫b0(exp(at)−exp(−at))exp(−st)dt

=limb→∞12∫b0exp((a−s)t)−exp((−a−s)t)dt

limb→∞12[(1a−sexp((a−s)b)−1−a−sexp((−a−s)b))−(1a−s−1−a−s)]

But taking the limit as b→∞ yields infinity terms.

REFER TO THE ATTACHMENT ALSO!!

Answer 2

Mate your answer is attached above .....