Laplace Transform of $(2\sinh(x))^2$

Question

I have to perform the Laplace Transform on the indicated function: $$f(t)=(2\sinh(t))^2$$

I have to perform the given operation:

$$\mathcal{L}\{f(t)\}$$

However I am limited to this theorem $7.1.1$.

Theorem $7.1.1$

\begin{align}&a) \ \mathcal{L}\{1\}=\frac{1}s \\ &b)\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}} \\ &c)\mathcal{L}\{e^{at}\}=\frac{1}{s-a}\\&d)\mathcal{L}\{\sin(kt)\}=\frac{k}{s^2+k^2}\\&e)\mathcal{L}\{\cos(kt)\}=\frac{s}{s^2+k^s} \\&f)\mathcal{L}\{\sinh(kt)\}=\frac{k}{s^2-k^2}\\&g)\mathcal{L}\{\cosh(kt)\}=\frac{s}{s^2-k^2}\end{align} My question is how do this way I can do this way as well:

Definition 7.1.1

$$\displaystyle{\mathcal{L}\{f(t)\}=\int_\limits{0}^\infty e^{-st}f(t)dt}$$ The only thing that is tripping me up is arriving at the answer using those theorems otherwise I can get the answer using the definition. Thank you in advance.

Answer 1

If you know $\sinh(x)=\frac{e^x-e^{-x}}{2}$, then $(2\sinh(x))^2=4\sinh^2(x)=(e^x-e^{-x})^2=e^{2x}-2+e^{-2x}$. Then use the fact that Laplace transform is linear so $$ \mathcal{L}((2\sinh(x))^2)=\mathcal{L}(e^{2x})-2\mathcal{L}(1)+\mathcal{L}(e^{-2x}). $$

Answer 2

Hint: Use the fact that $$2\sinh(x) =e^x - e^{-x}$$

So, $$4\sinh^2(x) = e^{2x} -2 +e^{-2x}$$

Answer 3

$$\cosh (2t) = \cosh^2 t + \sinh^2 t$$ $$\cosh (2t) = 1 + 2\sinh^2 t$$ Therefore: $$\sinh^2 t = \frac 12 (\cosh (2t)-1)$$ $$\mathcal{L}(\sinh^2 t )= \frac 12 \mathcal{L}(\cosh (2t))-\dfrac 1{2s}$$ $$\mathcal{L}(\sinh^2 t )= \frac 12 \dfrac s{s^2-4}-\dfrac 1{2s}$$ $$\mathcal{L}(\sinh^2 t )= \dfrac 2{s(s^2-4)}$$