antidervative of $\ln\left|\tan{\left(\dfrac{\cos^{-1}{\left(\left|x\right|\right)}}{2} + \dfrac{\pi}{4} \right)}\right| $

Question

Question

How can the antiderivative of $$ \ln\left|\tan{\left(\dfrac{\cos^{-1}{\left(\left|x\right|\right)}}{2} + \dfrac{\pi}{4} \right)}\right| $$ be obtained?

Answer 1

A hint from trigonometry

Consider the Figure below, where $\triangle ABC$ is right-angled, $\overline{AB} = |x|$ and $\overline{AC}=1$, so that, by definition $$\alpha = \angle CAB = \arccos |x|.$$ Draw the bisector of $A$, that intersects $BC$ in $D$. From $D$ draw the perpendicular to $AD$. Let, on it, $E$ be a point such that $ED \cong AD$. Finally draw from $E$ the perpendicular to $AB$, that intersects $AB$ in $H$.

Clearly we have \begin{eqnarray} \angle EAB &=& \frac{\alpha}2+\frac{\pi}4=\\ &=&\frac{\arccos |x|}2 + \frac{\pi}4, \end{eqnarray} and \begin{eqnarray} \tan \angle EAB &=& \tan \left(\frac{\arccos |x|}2 + \frac{\pi}4\right)=\\ &=& \frac{\overline{EH}}{\overline{AH}}. \end{eqnarray}

So our first aim is to write this ratio in terms of $x$ "directly".

First we calculate $$\overline{AD} = \frac{\overline{AB}}{\cos\frac{\alpha}2}=\frac{\sqrt 2 |x|}{\sqrt{1+|x|}},$$ by the bisection formula, and

$$\overline{BD} = |x|\sqrt{\frac{1-|x|}{1+|x|}},$$ by Pythagorean Theorem on $\triangle ABD$.

Let then $F$ be the intersection between $EH$ and $AD$. We have $\triangle EFD\sim\triangle ABD$, and, therefore $$\overline{EF} = \frac{2|x|}{1+|x|}.$$

Pythagorean Theorem on $\triangle AFD$ gives $$\overline{FD} = \frac{|x|\sqrt{2(1-|x|)}}{1+|x|}.$$

As a consequence $$\overline{AF} = \frac{\sqrt 2 |x| (\sqrt{1+|x|}-\sqrt{1-|x|})}{1+|x|}.$$

Similarity $\triangle AFH\sim\triangle ABD$ yields

$$\overline{FH} = \frac{|x|\sqrt{1-|x|}(\sqrt{1+|x|}-\sqrt{1-|x|})}{1+|x|}$$ and $$\overline{AH} = \frac{|x|(\sqrt{1+|x|}-\sqrt{1-|x|})}{\sqrt{1+|x|}}.$$

The tangent we are looking for is thus \begin{eqnarray} \tan \angle EAB &=& \frac{\frac{2|x|}{1+|x|}+\frac{|x|\sqrt{1-|x|}(\sqrt{1+|x|}-\sqrt{1-|x|})}{1+|x|}}{\frac{|x|(\sqrt{1+|x|}-\sqrt{1-|x|})}{\sqrt{1+|x|}}} =\\ &=&\frac{1+|x|+\sqrt{1-x^2}}{1+|x|-\sqrt{1-x^2}}=\\ &=&\frac{\left(1+|x|+\sqrt{1-x^2}\right)^2}{(1+|x|)^2-1+x^2}=\\ &=&\frac{1+\sqrt{1-x^2}}{|x|}. \end{eqnarray}

Therefore, integrating by parts leads to the result

\begin{eqnarray} \mathcal I &=& \int\log\left|\tan\left(\frac{\arccos x}2+\frac{\pi}4\right)\right| dx=\\ &=& \int \log\left|\frac{1+\sqrt{1-x^2}}{|x|}\right| dx=\\ &=& \int \log\left(\frac{1+\sqrt{1-x^2}}{|x|}\right) dx=\\ &=& \boxed{x\log\left(\frac{1+\sqrt{1-x^2}}{|x|}\right) + \arcsin x + C}. \end{eqnarray}