7
$\begingroup$

I am very confused by this. I know that the derivative of $\text{arcsec}(x)$ is $\dfrac{1}{|x|\sqrt{x^2-1}}$. However, if you plug in the integral of $\dfrac{1}{x\sqrt{x^2-1}}$ into wolfram alpha it gives some other answer with an inverse tangent: $$ \int \dfrac{1}{x\sqrt{x^2-1}}dx = - \tan^{-1}\Bigg(\frac{1}{\sqrt{x^2-1}} \Bigg) +C $$

I was just wondering why this is, or why wolfram is giving something totally different. Are they equivalent?

$\endgroup$
2
  • $\begingroup$ Wolfram Alpha is very very rarely wrong... $\endgroup$
    – user65203
    May 6 '19 at 14:20
  • $\begingroup$ In case of doubt, take the derivative. $\endgroup$
    – user65203
    May 6 '19 at 14:21
5
$\begingroup$

The two answers are equivalent. Remember that $\sin$ and $\tan$, from a trigonometric point of view, are just ratios of sides of a right angled triangle. The $\tfrac{1}{\sqrt{x^2-1}}$ in the wolfram result looks suspiciously like an application of Pythagoras's theorem, doesn't it? You can do the calculation yourself.

$\endgroup$
2
  • 2
    $\begingroup$ Why the complicated form though? Is this in some way more general? $\endgroup$
    – Aritra Das
    Apr 13 '16 at 20:05
  • $\begingroup$ Imagine a right-angled triangle with an angle $\theta$, hypotenuse $1$, and opposite side $x$. So what is $\sin\theta$? It's simply $x$. It escalates quickly. What about $\tan\theta$? By Pythagorus's Theorem, the adjacent side is given by $\sqrt{1-x^2}$. Hence $\tan\theta$ = \frac{x}{\sqrt{1-x^2}}. This kind of calculation is general, if you assign $x$ to one side and $1$ to another. $\endgroup$ Apr 13 '16 at 20:50

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.