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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\ $ ?

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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.

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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 $[0,\infty).)$

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  • $\begingroup$ One could make $\dagger$ more elementary: For the substitution one has to select a domain of $u$ where the chosen reparametrization function is (of course defined and) strictly monotonous. Thus the split at the symmetry axis $u=0$ of the cosine or secans. $\endgroup$ Sep 12 at 8:56
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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, $$\int_{-1}^x\sqrt{t^2-1}dt=\frac x2\sqrt{x^2-1}+\cosh^{-1}(x)$$

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