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.

  • $\begingroup$ 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

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:

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

  • 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

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.


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 $2\pi-9/5 -(2\pi-8/5)=-1/5.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.