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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

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  • $\begingroup$ 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. $\endgroup$ – Sangchul Lee May 6 '18 at 12:27
  • $\begingroup$ @SangchulLee If I am correct, then why is the split in domains not reflected in the final answer? (See my edit of the question) $\endgroup$ – GambitSquared May 6 '18 at 13:04
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    $\begingroup$ 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 $\lvert\tan\theta\rvert^3=-\tan^3\theta$. $\endgroup$ – Sangchul Lee May 6 '18 at 13:15
  • $\begingroup$ @SangchulLee But according to wolfram alpha we don't need the $sign(x)$ $\endgroup$ – GambitSquared May 6 '18 at 13:16
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    $\begingroup$ 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 $$\int\frac{dx}{x^2(x^2-1)^{3/2}}=\mathrm{sign}(\tan\theta)\int\frac{d\theta}{\sec\theta\tan^2\theta}=\frac{1-2\sec^2 \theta}{\sec\theta\lvert \tan\theta\rvert}+C=\frac{1-2x^2}{x\sqrt{x^2-1}}+C.$$ $\endgroup$ – Sangchul Lee May 6 '18 at 14:48
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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.


Incidentally, a point often overlooked in introductory courses is that when you antidifferentiate a function on a disconnected domain, each component of the domain gets is own distinct constant of integration.

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.

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  • $\begingroup$ Thanks for an interesting alternative to my method. But it doesn't really address the question if and why we can remove the absolute sign in that case... Also see the edit to the question, which shows that the solution is not divided into seperate domains. $\endgroup$ – GambitSquared May 6 '18 at 13:07

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