# Integral of $\frac{1}{x\sqrt{x^2-1}}$

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?

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

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.
• 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. Apr 13 '16 at 20:50