# How to find Inverse Laplace transform of $\tan^{-1}\left(\frac{2}{s^2}\right)$?

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## Solution 1

You can write $$\tan ^{-1}\left(\frac{2}{s^2}\right)=\tan ^{-1}\left(\frac{1}{s-1}\right)-\tan ^{-1}\left(\frac{1}{s+1}\right)$$ $$\mathcal{L}_s^{-1}\left[\tan ^{-1}\left(\frac{2}{s^2}\right)\right](t)=\mathcal{L}_s^{-1}\left[\tan ^{-1}\left(\frac{1}{s-1}\right)\right](t)-\mathcal{L}_s^{-1}\left[\tan ^{-1}\left(\frac{1}{s+1}\right)\right](t)$$

Now remember the property $\mathcal{L}_s^{-1}[f(s)](t)=-\dfrac{\mathcal{L}_s^{-1}\left[f'(s)\right](t)}{t}$

Derivative of $\tan ^{-1}\left(\dfrac{1}{s-1}\right)$ is

$-\dfrac{1}{(s-1)^2+1}$ and any Laplace transform table can tell that

$\mathcal{L}_s^{-1}\left[-\dfrac{1}{(s-1)^2+1}\right](t)=-e^t \sin t$

Therefore $\mathcal{L}_s^{-1}\left[\tan ^{-1}\left(\dfrac{1}{s-1}\right)\right](t)=-\dfrac{-e^t \sin t}{t}=\dfrac{e^t\,\sin t}{t}$

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 $$\mathcal{L}_s^{-1}\left[\tan ^{-1}\left(\frac{2}{s^2}\right)\right](t)=\dfrac{e^t\,\sin t}{t}-\dfrac{e^{-t}\, \sin (t)}{t}=\dfrac{\left(e^t-e^{-t}\right)\sin t}{t}=\frac{2\sinh t \sin t}{t}$$

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 : $n=1$

$$-\frac{\mathcal{L}_s^{-1}\left[\frac{\partial }{\partial s}\tan ^{-1}\left(\frac{2}{s^2}\right)\right](t)}{t}=-\frac{\mathcal{L}_s^{-1}\left[-\frac{4 s}{4+s^4}\right](t)}{t}=-\frac{\mathcal{L}_s^{-1}\left[-\frac{4 s}{\left(2-2 s+s^2\right) \left(2+2 s+s^2\right)}\right](t)}{t}=-\frac{\mathcal{L}_s^{-1}\left[-\frac{1}{2-2 s+s^2}+\frac{1}{2+2 s+s^2}\right](t)}{t}=-\frac{\mathcal{L}_s^{-1}\left[-\frac{1}{((1-i)-s) ((1+i)-s)}+\frac{1}{((1-i)+s) ((1+i)+s)}\right](t)}{t}=-\frac{\mathcal{L}_s^{-1}\left[\frac{i}{2 ((-1-i)+s)}-\frac{i}{2 ((-1+i)+s)}-\frac{i}{2 ((1-i)+s)}+\frac{i}{2 ((1+i)+s)}\right](t)}{t}=-\frac{i e^{(-1-i) t} \left(-1+e^{2 i t}\right) \left(-1+e^{2 t}\right)}{2 t}=\frac{2 \sin (t) \sinh (t)}{t}$$

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### Andrew Fount

Updated on August 01, 2022

How to find Inverse Laplace transform of $\tan^{-1}\left(\frac{2}{s^2}\right)$?