Inverse Trig Integration Resulting in $\sec^{-1}|x|$

Question

$\int \frac{1}{x\sqrt{{x^2}+1}} dx =\sec^{-1} |x| + C$

I'm a bit confused as to why the solution is $\sec^{-1}|x|$ instead of $\sec^{-1}x$, here is what I did:

let $x = \sec\theta, dx = \sec\theta\cdot \tan\theta \cdot d\theta$

$\int \frac{1}{x\sqrt{{x^2}+1}} dx = \int \frac{\sec\theta\cdot \tan\theta}{\sec\theta \sqrt{{\sec^2}\theta - 1}}dθ = \int \frac{\tan\theta}{\sqrt{\tan^2 \theta}}d\theta = \int1 \cdot d\theta = \theta + C = \sec^{-1}x + C $

Why is it $\sec^{-1}|x|$ instead of $\sec^{-1}x$?

Thanks in advance, and sorry if this question has been posted before.

It's $x\sqrt{x^2+1}$ or is it $x\sqrt{x^2-1}$ in the denominator? — Maadhav, Dec 06 '17 at 05:31

score 1 · Answer 1 · answered Dec 06 '17 at 05:21

1

For real $a,$ $$\sqrt{a^2}=|a|$$

$=+a$ if $a\ge0$

$=-a$ otherwise

Now use https://en.m.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values

answered Dec 06 '17 at 05:21

lab bhattacharjee

274,582

See also: https://math.stackexchange.com/questions/1224415/how-do-i-prove-that-arccosx-arccos-x-pi-when-x-in-1-1 – lab bhattacharjee Dec 06 '17 at 05:27

Inverse Trig Integration Resulting in $\sec^{-1}|x|$

1 Answers1