Laplace Transform of $\cosh^2(3t)$

Could someone help me on laplace transfrom ?

Using Laplace transform of derivative of $f(t)$,

Find the Laplace transform for

A) $\cosh^2(3t)$

How to derive it using Laplace transform of derivative ??

Then answer given is $\displaystyle\frac{S^2-18}{S(S-6)(S+6)}$

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even if you write the integral explicitly, you'll have a bunch of exponentials which are extremely easy to integrate, isn't it? –  Ilya Jan 17 '13 at 9:22
I completely agree with Ilya. You have only three pure exponential terms to integrate after expanding. –  Sangchul Lee Jan 17 '13 at 9:28
no you cant because question asking to use the method of Laplace transform of derivative –  Garett Jan 17 '13 at 9:28

You can do it as follows:

1- Take @Ilya's suggestion to write $\cosh(3t)$ as:

$$\left(\frac{1}{2}\exp(3t)+\frac{1}2\exp(-3t)\right)^2=\frac{1}4\exp(6t)+\frac{1}2+\frac{1}4\exp(-6t)$$

Or

2- Use the fact that: $$\cosh^2(3t)=\frac{1+\cosh(6t)}{2}$$

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thanks man. I managed to solve the problem –  Garett Jan 20 '13 at 10:42
+1 Nice work! ;-) –  amWhy Feb 15 '13 at 0:03