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in my math class we were given a list of indefinite integrals, and one of them was:

$$\int \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}}$$

My working:

$$\int \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}}=\int \frac{dx}{(x+2)\sqrt{(x+2)^2-1}}$$

Then I used the substitution $x+2=\sec t$ to get:

$$\int \frac{\tan t}{\sqrt{\sec^2 t-1}}dt=\int \frac{\tan t}{|\tan t|}dt= t\,\text{sgn}\, (\tan t)+C...$$

Then I checked the answer sheet, and this is what they did:

$$\int \frac{\tan t}{\sqrt{\sec^2 t-1}}dt=\int dt=t+C=\text{arcsec}(x+2)+C$$

What I don't understand is, why are they allowed to say $\sqrt{\sec^2 t-1}=\tan t?$ I tried to put some values in and I have found that:

$$\int_{\sec \left(\frac{8}{5}\right)-2}^{\sec \left(\frac{9}{5}\right)-2} \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}}<0$$ but according to the answer sheet I would get $\dfrac{1}{5}$

My answer looks wrong, I would be happy if someone could explain what the problem is, and also why we are allowed to simplify like they did.

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  • $\begingroup$ There is an identity that "converts" the radical into a tangent. Write the -1 as -cos²/cos² and use Pyth. identity $\endgroup$ – imranfat May 10 '13 at 15:37
  • $\begingroup$ @imranfat that's not my question, I am asking why they can write $\sqrt{\tan^2 t}=\tan t$ instead of $\sqrt{\tan^2 t}=|\tan t|$ $\endgroup$ – Little miss sunshine May 10 '13 at 15:46
  • $\begingroup$ @Littlemisssunshine (what a name, first of all!) You see the substitution you made for $t$: $x + 2 = sec(t)$, if you modify it a bit, you get $ t = arcsec(x + 2) $. Now the principal range of $arcsec$ is $ [0, \pi] $, and hence $t$ must be positive. So $|\tan t| = \tan t$. $\endgroup$ – Parth Thakkar May 10 '13 at 16:23
  • $\begingroup$ Btw, I do know that this isn't exactly an answer (and hence was posted as a comment). There's a flaw: $ x + 2 = sec(t) => t = arcsec(x + 2) $ only if $t$ lies in the principle range. So, it is just beating around the bush. But so far, I've seen in solutions to integration problems, that such things aren't really bothered about. I maybe horribly wrong about this, I don't know. $\endgroup$ – Parth Thakkar May 10 '13 at 16:28
  • $\begingroup$ Ah, the absolute value. Parth, usually it isn't important, the domain of the original integral is as such that it "converts" in such a way that the abs. value becomes redundant. But now I am in for a counterexample which I am going to search for, I guess. (Now I do know with finding a limit, that one has to be careful with the use of absolute values) $\endgroup$ – imranfat May 10 '13 at 16:36
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The answer of the book looks wrong. Since the principal range of $\sec x$ is:

  • $\left[ 0; \frac{\pi}{2} \right)$ if $x \ge 1$, and on this domain, $\tan t$ is positive.
  • $\left( \frac{\pi}{2} ; \pi \right]$ if $x \le -1$, whereas on this domain, $\tan t$ is negative.

So, in fact, the solution to that integral should be split in to 2 different parts:

$$\int\limits \frac{dx}{(x+2)\sqrt{(x+1)(x+3)}} = \left\{ \begin{array}{ll} -\mbox{arcsec}(x + 2) + C_1 &, \mbox{for }x < -3 \\ \mbox{arcsec}(x + 2) + C_2 &, \mbox{for }x > -1 \end{array} \right.$$

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  • 2
    $\begingroup$ Another approach: Let $u=x+2$, we need to integrate $du/u\sqrt{u^2-1}$. Set $v=\sqrt{u^2-1}$, we have $du/u=vdv/(v^2+1)$ and $dx/(x+2)\sqrt{(x+1)(x+3)}=dv/(v^2+1)$, therefore the answer is $\arctan v$ where $v=\sqrt{(x+1)(x+3)}$. $\endgroup$ – Yai0Phah May 10 '13 at 17:50
  • $\begingroup$ @user49685 I think you're right, there must be 2 parts of the solution! $\endgroup$ – Parth Thakkar May 11 '13 at 5:10
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I'm sure that you're more rigorous than the book. But actually it is unnecessary when encountered in definite integration. You just need to be aware of that $ \int \frac{1}{x \sqrt{x^2-1}} dx = {\rm{arcsec}} (x) + C $ is only valid when x>1. Just like in the example of $ y = \ln (x) $, we normally have $ \int \frac{1}{x} dx = \ln (x) $ where you know that x must be positive. When the lower and up limit are negative in definite integration, we just need to convert them to positve region.

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I know this is an old question, but I think I know why your answer sheet might have ignored the absolute value. Some calculus books (e.g. Stewart´s) define the arcsec function as the inverse function of the secant restricted to $[0,\pi/2)\cup[\pi,3\pi/2)$, instead of the usual $[0,\pi/2)\cup(\pi/2,\pi]$. I, personally, think this is a bad idea. However, by doing this, you avoid the absolute value "problem" in these integrals. Note that now the tangent is non-negative in the range of the arcsec. But then you have to be careful, because with this definition we actually have: $$arcsec(sec(8/5))=2\pi-8/5$$ and similarly with $9/5$, so that the integral in your example would give $2\pi-9/5 -(2\pi-8/5)=-1/5.$

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