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Math Scrap 1

Considering:

∫sech⁡ax dx=1atan⁡−1(sinh⁡ax)=2atan⁡−1(tanh⁡ax2) \int{\sech{ax} \,\diff{x}} = \frac{1}{a} \tan^{-1}{\parens{\sinh{ax}}} = \frac{2}{a} \tan^{-1}{\parens{\tanh{\frac{ax}{2}}}}

Prove that:

1atan⁡−1(sinh⁡ax)=2atan⁡−1(tanh⁡ax2) \frac{1}{a} \tan^{-1}{\parens{\sinh{ax}}} = \frac{2}{a} \tan^{-1}{\parens{\tanh{\frac{ax}{2}}}}

Let us use the following identities for this proof:

tan⁡−1x=2tan⁡−1(x1+1+x2)sinh⁡x=ex−e−x2=e2x−12extanh⁡x=ex−e−xex+e−x=e2x−1e2x+1\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=1atan⁡−1(sinh⁡ax)=2atan⁡−1(sinh⁡ax1+1+sinh⁡2ax) \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⁡ax2=sinh⁡ax1+1+sinh⁡2ax \tanh{\frac{ax}{2}} = \frac{\sinh{ax}}{1 + \sqrt{1 + \sinh^2{ax}}}

To simplify further working, we will change the variable:

tanh⁡x=sinh⁡2x1+1+sinh⁡22x \tanh{x} = \frac{\sinh{2x}}{1 + \sqrt{1 + \sinh^2{2x}}}

Starting with the RHS:

RHS=e4x−12e2x1+1+(e4x−12e2x)2=e4x−12e2x(1+1+e8x−2e4x+14e4x)=e4x−12e2x(1+e8x−2e4x+1+4e4x4e4x)=e4x−12e2x(1+e8x+2e4x+14e4x)=e4x−12e2x(1+(e4x+1)24e4x)=e4x−12e2x(1+e4x+12e2x)=e4x−12e2x+e4x+1=(e2x−1)(e2x+1)(e2x+1)2=e2x−1e2x+1=tanh⁡x=LHS\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.