Solve using trig substitution:


  1. $u = x = \left({3\sec\theta}\right)^{2} = 9\sec^{2}\theta$ and $ dx = \sec\theta\tan\theta$
  2. $\theta = \sec^{-1}\left(\frac{x}{3}\right)$ and $ \sqrt{x^2-9} = 3\tan\theta$
  3. $\int \frac{\sec\theta\tan\theta}{9\sec\theta\tan\theta}d\theta$
  4. $\frac{1}{9}\int d\theta = \frac{1}{9}\theta = \frac{1}{9}\sec^{-1}\left(\frac{x}{3}\right) + C$

As you can see, we end up at $$ \bbox[5px,border:2px solid #282] {\frac{1}{9}\sec^{-1}\left(\frac{x}{3}\right) + C} $$

However, when I use calculators to solve the same problem, I end up getting $$ \bbox[5px,border:2px solid red] {\frac{1}{3}\tan^{-1}\left(\frac{\sqrt{x^2-9}}{3}\right) + C} $$ or $$ \bbox[5px,border:2px solid red] {\sqrt{x^2-9} + 3tan^{-1}\left(\frac{3}{\sqrt{x^2-9}}\right) + C} $$

WolframAlpha does not explain why (obviously I don't know) $ \sec^{-1}x$ ended up being simplified to $\tan^{-1}x$, the $ x $ parameter being whatever with substitute in for.

My guess is it has to do with the trig substitution rules.

I haven't found any trig identity that looks like that or why the answer is simplified this way, and differently if that matters at all.

Can someone explain why?Thanks.

  • $\begingroup$ You made a mistake in 1. You should have found $dx=3\sec\theta\tan\theta$, not $dx=\sec\theta\tan\theta$. As a result, the green box should be ${1\over3}\sec^{-1}\left(x\over3\right)+C$. $\endgroup$ – Barry Cipra May 4 '17 at 8:45
  • $\begingroup$ A comment on a very minor formatting mistake: You forgot the "\" from "\arctan" the final call. The edit system needs at least 6 changes so I was unable to do it. Nothing at all wrong with what you've posted mathematically, just as again a very minor latex formatting issue. $\endgroup$ – user150203 Dec 4 '18 at 23:53

Try making a triangle. with hypotenuse $x$, base $3$. You will get the height to be $\sqrt{x^2-9}$. So, wrt to that angle, $$\sec^{-1}(\frac{x}{3})=\tan^{-1}\frac{\sqrt{x^2-9}}{3}$$

  • $\begingroup$ This is the only answer that makes sense. I can see why now. Thanks. $\endgroup$ – TheRealChx101 Jul 5 '17 at 14:13


$$\cos^2t+\sin^2t=1$$ you draw


$$\tan t=\sqrt{\sec^2t-1}.$$ Then with $t=\text{arcsec }x$, taking the arc tangent,

$$t=\arctan\sqrt{x^2-1}=\text{arcsec }x.$$


$$\text{arcsec}\sqrt{x^2+1}=\arctan x.$$


Let us prove that : $$\sec^{-1}\left(\frac{x}{3}\right)=\tan^{-1}\left(\frac{\sqrt{x^2-9}}{3}\right)$$

Starting from L.H.S., put $x=3 \sec t$, you'll get :


Now do the same with R.H.S.

$$R.H.S.= \tan^{-1}\left(\frac{\sqrt{x^2-9}}{3}\right)=\tan^{-1}\left( \sqrt{\sec^2t-1 }\right)=\tan^{-1}({\tan t})=t$$

Therefore $$L.H.S=R.H.S$$

  • $\begingroup$ You need to take care with domains and ranges. The arctan on the RHS always returns a value between $0$ and $\pi/2$, but if $x\le-3$, the arcsec on the LHS returns a value between $\pi/2$ and $\pi$. The "$+C$" plays a role! $\endgroup$ – Barry Cipra May 4 '17 at 8:51

$\sec^{-1}(x/3) = \alpha \\ \therefore \sec(\alpha) = 1/\cos(\alpha) = x/3 => \cos(\alpha) = \dfrac{3}{x}$


$\therefore \tan^{-1}(\tan(\alpha))=\alpha=\sec^{-1}(x/3)=\tan^{-1}(\dfrac{\sqrt{x^2-9}}{3})$

  • $\begingroup$ Of course this is only true if $x>0$. $\endgroup$ – Soroush khoubyarian May 4 '17 at 8:54

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