sech Integral

Let us use the following identities for this proof:

$$\begin{gather*} \tan^{-1}{x} = 2 \tan^{-1}{\parens{\frac{x}{1 + \sqrt{1 + x^2}}}} \\ \sinh{x} = \frac{e^x - e^{-x}}{2} = \frac{e^{2x} - 1}{2 e^x} \\ \tanh{x} = \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{e^{2x} - 1}{e^{2x} + 1} \end{gather*}$$

We start by rewriting LHS:

$$\lhs = \frac{1}{a} \tan^{-1}{\parens{\sinh{ax}}} = \frac{2}{a} \tan^{-1}{\parens{\frac{\sinh{ax}}{1 + \sqrt{1 + \sinh^2{ax}}}}}$$

Now, to finish the proof, we will now show the following to be true:

$$\tanh{\frac{ax}{2}} = \frac{\sinh{ax}}{1 + \sqrt{1 + \sinh^2{ax}}}$$

To simplify further working, we will change the variable:

$$\tanh{x} = \frac{\sinh{2x}}{1 + \sqrt{1 + \sinh^2{2x}}}$$

Starting with the RHS:

$$\begin{align*} \rhs &= \frac{\frac{e^{4x} - 1}{2 e^{2x}}}{1 + \sqrt{1 + \parens{\frac{e^{4x} - 1}{2 e^{2x}}}^2}} = \frac{e^{4x} - 1}{2 e^{2x} \parens{1 + \sqrt{1 + \frac{e^{8x} - 2 e^{4x} + 1}{4 e^{4x}}}}} = \frac{e^{4x} - 1}{2 e^{2x} \parens{1 + \sqrt{\frac{e^{8x} - 2 e^{4x} + 1 + 4 e^{4x}}{4 e^{4x}}}}} = \frac{e^{4x} - 1}{2 e^{2x} \parens{1 + \sqrt{\frac{e^{8x} + 2 e^{4x} + 1}{4 e^{4x}}}}} \\ &= \frac{e^{4x} - 1}{2 e^{2x} \parens{1 + \sqrt{\frac{\parens{e^{4x} + 1}^2}{4 e^{4x}}}}} = \frac{e^{4x} - 1}{2 e^{2x} \parens{1 + \frac{e^{4x} + 1}{2 e^{2x}}}} = \frac{e^{4x} - 1}{2 e^{2x} + e^{4x} + 1} = \frac{\parens{e^{2x} - 1} \parens{e^{2x} + 1}}{\parens{e^{2x} + 1}^2} = \frac{e^{2x} - 1}{e^{2x} + 1} = \tanh{x} = \lhs \end{align*}$$

as required.