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A question in my textbook says to evaluate $\displaystyle \int \frac{1}{\sqrt{x^2-a^2}}~dx$ where $a \gt 0$. I know how to solve the integral using trig substitution but what i do not understand is why is $a \gt 0$ necessary? I could solve it the same way even if $a \lt 0$ ?

Robert Shore
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wowf momf
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  • Yes indeed you could. – Anne Bauval Feb 02 '23 at 23:24
  • For some basic information about writing mathematics at this site see, e.g., here, here, here and here. – Another User Feb 02 '23 at 23:24
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    Note that since $a$ is squared, it being positive or negative can't affect the outcome. Assuming $a > 0$ is for your convenience, so you can write $\sqrt{a^2} = a$ instead of $\sqrt{a^2} = |a|$. – eyeballfrog Feb 02 '23 at 23:25
  • Anne Bauval which confuses me why would the textbook give an unnecessary piece of info?(James Stewart chp 7.3 Example 5. – wowf momf Feb 02 '23 at 23:26
  • Eyeballfrog but we would still have to deal with |tan n| when we do the trigonometric substitution. – wowf momf Feb 02 '23 at 23:35
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    I think Stewart is trying to help you by providing information that leads to an answer involving just plain $a$. Your careful reading made his attempt confusing rather than helpful. – Ethan Bolker Feb 03 '23 at 00:39
  • My guess is that the textbook wants you to consider the solution for $x>a$ which makes more sense if $a$ is positive and didn't want to complicate with $|x|>|a|$ which splits in two disjoint intervals with two different +C to consider (though perfectly fine too). I think it's just to make the exercise more accessible. – zwim Feb 03 '23 at 01:32

2 Answers2

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It is independent of the sign of $a$ since it's getting squared and even the solution of this integral contains $a^2$. So, the sign of $a$ doesn't really matter. $$\int\frac{dx}{\sqrt{x^2-a^2}}=\log\Big|x+\sqrt{x^2-a^2}\Big|+c$$

Hope this helps!!

Adil Ahamed
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Adding to E Bolker's above comment: since $(x^2-(-7)^2)$ immediately equals $(x^2-7^2),$ choosing Set 2 is mainly just to avoid the absolute value symbol. The formulae in Set 1 already require $a\ne0,$ and going for $a>0$ instead isn't a big ask.

  1. $\displaystyle \int \frac{\mathrm dx}{\sqrt{x^2-a^2}}=\operatorname{arccosh}\left(\frac x{|a|}\right)\quad\text{for } a\ne0,\,\;x\in(a,\infty) \\ \displaystyle \int \frac{\mathrm dx}{\sqrt{a^2-x^2}}=\operatorname{arcsin}\left(\frac x{|a|}\right)\quad\text{for } a\ne0\\ \displaystyle \int \frac{\mathrm dx}{\sqrt{x^2+a^2}}=\operatorname{arcsinh}\left(\frac x{|a|}\right)\quad\text{for } a\ne0$

  2. $\displaystyle \int \frac{\mathrm dx}{\sqrt{x^2-a^2}}=\operatorname{arccosh}\left(\frac x{a}\right)\quad\text{for } a>0,\,\;x\in(a,\infty) \\ \displaystyle \int \frac{\mathrm dx}{\sqrt{a^2-x^2}}=\operatorname{arcsin}\left(\frac x{a}\right)\quad\text{for } a>0\\ \displaystyle \int \frac{\mathrm dx}{\sqrt{x^2+a^2}}=\operatorname{arcsinh}\left(\frac x{a}\right)\quad\text{for } a>0$

ryang
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