Originally Posted by

**disclaimer** Thanks. It's just that using $\displaystyle arc\sec$ or $\displaystyle \sec$ is very uncommon in my neck of the woods. So I wanted to get something connected with $\displaystyle \sin$. Here's my reasoning:

$\displaystyle \int{\frac{dx}{x\sqrt{x^2-a^2}}}$

Let $\displaystyle \frac{1}{x}=t$

So $\displaystyle x=\frac{1}{t}\Longrightarrow{dx=-\frac{1}{t^2}dt}$

That way we get $\displaystyle \int{\frac{-t}{t^2\sqrt{\frac{1}{t^2}-a^2}}}dt$ which yields $\displaystyle -\int{\frac{dt}{\sqrt{1-a^2t^2}}}$

Now we can write it as $\displaystyle -\frac{1}{a}\int{\frac{dt}{\sqrt{\frac{1}{a^2}-t^2}}}$ and use the formula $\displaystyle \int{\frac{dx}{\sqrt{a^2-x^2}}}=\arcsin{\frac{x}{|a|}}+C$

That said, $\displaystyle -\frac{1}{a}\int{\frac{dt}{\sqrt{\frac{1}{a^2}-t^2}}}=-\frac{1}{a}\arcsin{at}+C$

And, finally coming back to our substitution, we get:

$\displaystyle \boxed{\int{\frac{dx}{x\sqrt{x^2-a^2}}}=-\frac{1}{a}\arcsin{\frac{a}{x}}+C}$

What do you think about it? I'm not sure about the absolute values, to my mind it should be $\displaystyle

-\frac{1}{|a|}\arcsin{\frac{|a|}{x}}+C

$.