How to find Inverse Laplace transform of $\tan^{1}\left(\frac{2}{s^2}\right)$?
Solution 1
You can write $$\tan ^{1}\left(\frac{2}{s^2}\right)=\tan ^{1}\left(\frac{1}{s1}\right)\tan ^{1}\left(\frac{1}{s+1}\right)$$ $$\mathcal{L}_s^{1}\left[\tan ^{1}\left(\frac{2}{s^2}\right)\right](t)=\mathcal{L}_s^{1}\left[\tan ^{1}\left(\frac{1}{s1}\right)\right](t)\mathcal{L}_s^{1}\left[\tan ^{1}\left(\frac{1}{s+1}\right)\right](t)$$
Now remember the property $\mathcal{L}_s^{1}[f(s)](t)=\dfrac{\mathcal{L}_s^{1}\left[f'(s)\right](t)}{t}$
Derivative of $\tan ^{1}\left(\dfrac{1}{s1}\right)$ is
$\dfrac{1}{(s1)^2+1}$ and any Laplace transform table can tell that
$\mathcal{L}_s^{1}\left[\dfrac{1}{(s1)^2+1}\right](t)=e^t \sin t$
Therefore $\mathcal{L}_s^{1}\left[\tan ^{1}\left(\dfrac{1}{s1}\right)\right](t)=\dfrac{e^t \sin t}{t}=\dfrac{e^t\,\sin t}{t}$
and in a similar way
$\mathcal{L}_s^{1}\left[\tan ^{1}\left(\frac{1}{s+1}\right)\right](t)=\dfrac{e^{t}\, \sin (t)}{t}$
so that $$\mathcal{L}_s^{1}\left[\tan ^{1}\left(\frac{2}{s^2}\right)\right](t)=\dfrac{e^t\,\sin t}{t}\dfrac{e^{t}\, \sin (t)}{t}=\dfrac{\left(e^te^{t}\right)\sin t}{t}=\frac{2\sinh t \sin t}{t}$$
Hope this helps
Solution 2
Using Inverse Laplace transform of derivatives theorem : $$\mathcal{L}_s^{1}\left[\frac{\partial ^nf(s)}{\partial s^n}\right](t)=(1)^n t^n F(t)$$
$$F(t)=\frac{\mathcal{L}_s^{1}\left[\frac{\partial ^nf(s)}{\partial s^n}\right](t)}{(1)^n t^n}$$ in my case : $n=1$
$$\frac{\mathcal{L}_s^{1}\left[\frac{\partial }{\partial s}\tan ^{1}\left(\frac{2}{s^2}\right)\right](t)}{t}=\frac{\mathcal{L}_s^{1}\left[\frac{4 s}{4+s^4}\right](t)}{t}=\frac{\mathcal{L}_s^{1}\left[\frac{4 s}{\left(22 s+s^2\right) \left(2+2 s+s^2\right)}\right](t)}{t}=\frac{\mathcal{L}_s^{1}\left[\frac{1}{22 s+s^2}+\frac{1}{2+2 s+s^2}\right](t)}{t}=\frac{\mathcal{L}_s^{1}\left[\frac{1}{((1i)s) ((1+i)s)}+\frac{1}{((1i)+s) ((1+i)+s)}\right](t)}{t}=\frac{\mathcal{L}_s^{1}\left[\frac{i}{2 ((1i)+s)}\frac{i}{2 ((1+i)+s)}\frac{i}{2 ((1i)+s)}+\frac{i}{2 ((1+i)+s)}\right](t)}{t}=\frac{i e^{(1i) t} \left(1+e^{2 i t}\right) \left(1+e^{2 t}\right)}{2 t}=\frac{2 \sin (t) \sinh (t)}{t}$$
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Andrew Fount
Updated on August 01, 2022Comments

Andrew Fount over 1 year
How to find Inverse Laplace transform of $\tan^{1}\left(\frac{2}{s^2}\right)$?
I don't understand how to rewrite it and use the table of Laplace transforms.

Andrew Fount about 6 yearsHow did you get the first equality?

Raffaele about 6 years@AndrewFount mat.utfsm.cl/scientia/archivos/vol11/Art2.pdf