In this question, I would like to investigate the location of the absolute value in the arcsecant integral.
Following this answer and this answer, we know the following is true:
$$
\frac{d}{dx}\sec^{-1}(x)=\frac{1}{|x|\sqrt{x^2-1}}.
$$
Taking indefinite integrals of both sides, we get
$$
\sec^{-1}(x)+C=\int \frac{1}{|x|\sqrt{x^2-1}} dx
$$
How can one use this result to deduce that
$$
\sec^{-1}(|x|)+C=\int \frac{1}{x\sqrt{x^2-1}} dx?
$$
(Notice that the absolute values used to be in the denominator of the right hand side, and now they are in the argument of the $\sec^{-1}$ function.)
What logical rule of deduction allows us to interchange the location of the absolute values on opposite sides of this equation?
Best Answer
Given that
$$ \sec^{-1}(u)+C=\int \frac{1}{|u|\sqrt{u^2-1}} du $$
let $u=|x|$. Then $du=\dfrac{|x|}{x}dx$ and
\begin{eqnarray} \sec^{-1}(|x|) +C&=&\int\frac{1}{|x|\sqrt{|x|^2-1}}\cdot \frac{|x|}{x}\,dx\\ &=& \int \frac{1}{x\sqrt{x^2-1}} dx? \end{eqnarray}