A question in my textbook says to evaluate $$\int \frac{dx}{\sqrt{x^2-a^2}}$$ where $a>0$.
I know how to solve the integral using trig substitution but what I do not understand is that my textbook uses the substitution $x = a\sec(t)$, $0<t<\pi/2$. When looking at this I realize that this does not account for negative values of $x$ or for $x=0$ since $\sec(t)$ in this interval is $>0$ what if the integral was given in the form of a proper integral from $-5$ to $5$? I read on a math forum that when $x<0$ we use the interval $\pi/2 <t < \pi$ but what about $x=0$?