In $\displaystyle \int \frac{1}{\sqrt{x^2-a^2}}~dx$ is it necessary for $a \gt 0$?

Question

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 $a \lt 0$ ?

Answer 1

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!!

Answer 2

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