Substituting $x = \sec(u)$ to evaluate $\int\sqrt{x^2-1}\ dx$: but why is $\tan(u) = \sqrt{x^2 - 1}$ rather than $\tan(u) = -\sqrt{x^2 - 1}\ $? I tried evaluating $\int\sqrt{x^2-1}\ dx$ using the substitution $x = \sec(u).$
I know my method gets to the same answer as wolframalpha, namely:
$\int\sqrt{x^2-1}\ dx = \frac{1}{2} \left(\ x \sqrt{x^2-1} - \ln\left|x + \sqrt{x^2-1}\right|\ \right) + C,\quad (*)$
but there is one step I can't justify.
When I got to $\int \tan^2(u)\sec(u)\ du = \frac{1}{2}\left(\ \tan(u)\sec(u) - \ln\left|\sec(u)+\tan(u)\right|\ \right) + C,$
I then have to substitute stuff back in in terms of $x$.
Now $x=\sec(u) \implies \tan^2(u) = x^2 - 1$. But I don't see how this implies $\tan(u) = \sqrt{x^2 - 1}$.
Comparing the graphs of $\sec(u)$ and $\tan(u)$, I don't see why not: $\ \tan(u) = -\sqrt{x^2 - 1}$, which would give:
$\int\sqrt{x^2-1}\ dx = \frac{1}{2} \left(\ -x \sqrt{x^2-1} - \ln\left|x - \sqrt{x^2-1}\right|\ \right) + C,\quad (**)$
which is a different answer than $(*)$ ?
Now I noticed that $(*) = -(**)\ $ (ignoring the $C \to -C)$.
I can see from the graph of $\sqrt{x^2-1}$ that for $x>1$, the definite integral $\int^x_1\sqrt{t^2-1}\ dt = (*),$ and for $x<-1,\ \int^{-1}_x\sqrt{t^2-1}\ dt = (**)$
So is the indefinite integral sort of poorly defined, or would you say it is:
$\int\sqrt{x^2-1}\ dx = \pm \frac{1}{2} \left(\ x \sqrt{x^2-1} - \ln\left|x + \sqrt{x^2-1}\right|\ \right) + C\ $ ?
 A: Let's differentiate and see by using which sign we get the integrand.
$\begin{align}\frac{d}{dx}\left[\frac{x}{2}\sqrt{x^2-1}-\frac12\ln|x+\sqrt{x^2-1}| + c\right]& = \frac{1}{2}\sqrt{x^2-1}+\frac x2\frac{x}{\sqrt{x^2-1}}-\frac{1+\frac{x}{\sqrt{x^2-1}}}{2(x+\sqrt{x^2-1})}\\
\\& = \frac{2x^2-1}{2\sqrt{x^2-1}}-\frac{1}{2\sqrt{x^2-1}}\\
\\& = \sqrt{x^2-1}\end{align}$
But using the negative sign yields $-\sqrt{x^2-1}$. So the one we've been using is correct.
A: The integration-by-substitution theorem applies here when the domain of the substitution function $\sec(u)$ is restricted† to $(0,\frac\pi2).$
(This substitution works even if the original domain of integration is, say, $[-2,-1]$.)
This is why only the positive output of $\tan(u)$ is taken.
   † due to the discontinuity of $\tan$ and $\sec$ at $\frac\pi2$ and the non-integrability of $\frac{\mathrm{d}}{\mathrm{d}x}\sec^{-1}$ on $[1,\infty)$ 
P.S. If you are acquainted with hyperbolic functions, $x=\cosh(u)$ is a more elegant substitution than $x=\sec(u).$ (As with the $\sec(u)$ substitution, there is an implicit understanding that the substitution function has a restricted domain—in this case $[0,\infty).)$
A: Another method to do this is, use hyperbolic functions.
Consider the change of variables $x=\cosh u$. Then for all $t]x\geq 1$
$$\int_1^x\sqrt{t^2-1}dt=\int_0^{\cosh^{-1}x}\sinh^2 u \;du$$
Since $\cosh(2u)=2\sinh^2u+1$, we have
$$\int_1^x\sqrt{t^2-1}dt=\frac12\int_0^{\cosh^{-1}x}\cosh(2u)-1 \;du=\frac14\sinh(2u)-\frac u2$$
Now,
$$\frac14\sinh(2u)-\frac u2=\frac12\left(\sinh u\cosh u-u\right)=\frac x2\sqrt{x^2-1}-\cosh^{-1}(x)$$
Hence,
$$\int_1^x\sqrt{t^2-1}dt=\frac x2\sqrt{x^2-1}-\cosh^{-1}(x)$$
Note that, for $x\geq 1$, $\cosh^{-1}x=\ln(x+\sqrt{x^2-1})$.
Similarly, for all $x\leq -1$ we have,
$$\int_{-1}^x\sqrt{t^2-1}dt=\frac x2\sqrt{x^2-1}+\cosh^{-1}(x)$$
