# 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\$$ ?

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.

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 $$[0,\infty).)$$

• 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. Sep 12 at 8:56

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, $$\int_{-1}^x\sqrt{t^2-1}dt=\frac x2\sqrt{x^2-1}+\cosh^{-1}(x)$$