# Why can we drop the absolute sign in this case of trig-substitution?

My question is about dropping the absolute-sign when solving this integral using substitution:

$$\int\frac{dx}{x^2(x^2-1)^{3/2}}$$

We can do the following substitution: $x=\sec\theta$ and $dx=\sec\theta\tan\theta d\theta$:

$$\int\frac{sec\theta\tan\theta d\theta}{\sec^2\theta(\tan^2\theta)^{3/2}}$$

Which equals:

$$\int\frac{\tan\theta d\theta}{\sec\theta|\tan\theta|^{3}}$$

Since $\theta=arcsec(x)$ we can conclude the following (I would think):

• When $x\ge1$ then $0\le\theta\le\frac{\pi}{2}$ so therefore $\tan\theta\ge 0$ and we can drop the absolute signs.
• But when $x\le-1$ then $\frac{\pi}{2}\le\theta\le\pi$ and then $\tan\theta\le 0$ and therefore I would think we cannot drop the absolute-sign.

But according to my calculus-book we can drop the the absolute sign. So what's going on here?

EDIT: What further confuses me is that the correct answer seems not to be split in two domains. The correct answer apparently is:

$$\frac{1-2x^2}{x\sqrt{x^2-1}}$$

... for all domains (!) according to Wolfram Alpha

• You are correct. Stripping off the absolute sign of $\lvert \tan\theta \rvert^3$ to obtain $\tan^3\theta$ is valid only when $\theta > 0$ (or equivalently, $x > 1$). Although not always the case, it seems that some calculus textbooks tend to be sloppy about specifying the correct domain on which a particular integration technique works. May 6, 2018 at 12:27
• @SangchulLee If I am correct, then why is the split in domains not reflected in the final answer? (See my edit of the question) May 6, 2018 at 13:04
• This substitution leads to $$\frac{1-2\sec^2 \theta}{\sec\theta \tan\theta}+C.$$ In order to plug $x=\sec\theta$, we still need to resolve the sign of $\tan\theta$, for otherwise you pickup the sign of $\sec\theta$ (which is equal to the sign of $x$) and obtain$$\frac{1-2\sec^2 \theta}{\sec\theta \tan\theta}=\operatorname{sign}(x)\frac{1-2x^2}{x\sqrt{x^2-1}}.$$ Of course, when $x > 1$ this sign just disappears and we can safely ignore this, but when you started from $x<1$, this sign also cancels out with the sign that you picked up from $\lvert\tan\theta\rvert^3=-\tan^3\theta$. May 6, 2018 at 13:15
• @SangchulLee But according to wolfram alpha we don't need the $sign(x)$ May 6, 2018 at 13:16
• Well, you picked up one $\mathrm{sign}(x)$ when stripping off the absolute signs from $\lvert\tan^3\theta\rvert$. But then you pick up another $\mathrm{sign}(x)$ when rewriting the integral $$\int\frac{d\theta}{\sec\theta\tan^2\theta}=\frac{1-2\sec^2\theta}{\sec\theta\tan\theta}+C$$ in terms of $x$. So they eventually cancel out in both cases $x>1$ and $x<1$. Putting altogether, we have $$\int\frac{dx}{x^2(x^2-1)^{3/2}}=\mathrm{sign}(\tan\theta)\int\frac{d\theta}{\sec\theta\tan^2\theta}=\frac{1-2\sec^2 \theta}{\sec\theta\lvert \tan\theta\rvert}+C=\frac{1-2x^2}{x\sqrt{x^2-1}}+C.$$ May 6, 2018 at 14:48

You can choose a different domain for $\theta$; for example, $\theta \in (0, \frac{\pi}{2}) \cup (\pi, \frac{3\pi}{2})$.
As $\theta$ ranges from $\pi$ to $3\pi/2$, $\sec(\theta)$ decreases from $-1$ to $-\infty$ and $\tan(\theta)$ increases from $0$ to $+\infty$.
In particular, on this domain for $\theta$ the substitution $x = \sec(\theta)$ still covers the domain $x \in (-\infty, -1)$, but you have $\tan(\theta)$ remaining positive.
So, if you want to speak in the full generality of covering both the $x > 1$ and $x < -1$ domains, you need to remember that each of those domains gets is own separate constant of integration.