Why do we ignore the absolute value in hyperbolic substitutions?

Question

For the integral $\displaystyle \int \frac{1}{\sqrt{x^{2}-1}} \,dx$, if we let $x= \cosh(y)$, then the root becomes $\sqrt{\sinh^2 (y)}$.

Why can one ignore the absolute value to give the answer of the root as $\sinh(y)$ only?

Answer 1

It is common practice to try substitutions without caring for "mundanities" such as sign discussions, and postpone them, the main challenge being to find an antiderivative.

In some cases, the domain of integration is such that no sign error is possible. In other cases, reverting to original variables may annihilate the sign problem. Fixing signs can also be done in the end, when checking the derivative. So there is no hurry.

In the present case, the integrand is an even function so that the antiderivative will be odd, and solving in the positives is enough (complete solution by mirroring).

Answer 2

The integrand function $ x\mapsto \frac{1}{\sqrt{x^2-1}} $ is defined at $ (-\infty,-1)\cup (1,+\infty)=I \cup J$.

If you look for the antiderivative at the interval $ J=(1,\infty) $, you will put $$x=\cosh(t)$$ with $ t>0$ and $\sinh(t)>0$.

But if you want the antiderivative at $ I=(-\infty,-1) $, you will need the substitution $$x=-\cosh(u)$$ with $ u>0 $ and $\sinh(u)>0$

Answer 3

One has to consider such substitutions on a case-by-case basis; I'll illustrate that by going beyond an implicit assumption in the problem to see what happens more broadly. I'll nonetheless assume throughout you only want to integrate over those real $x$ for which the integrand is also real.

For $x\ge1$, substitute $x=\cosh y$ with $y\ge0$${}^\dagger$: $\sqrt{x^2-1}$ is positive, so is $\sinh y$ as opposed to $-\sinh x$, whence the antiderivative's $x\ge1$ behaviour is of the form $\operatorname{arcosh}x+C_+$.

(${}^\dagger$ As @Snacc notes, this ensures an injective substitution.)

For $x\le-1$, substitute $x=-\cosh y$ with $y\ge0$: $\sqrt{x^2-1}$ is positive, so is $\sinh y$ as opposed to $-\sinh y$, whence the antiderivative's $x\le-1$ behaviour is of the form $-\operatorname{arcosh}(-x)+C_-$.

Therefore, the most general antiderivative on $\Bbb R\setminus(-1,\,1)$ is $\operatorname{sgn}x\operatorname{arcosh}|x|$ plus a locally constant function which can have different constant values $C_\pm$ either side of $(-1,\,1)$. We usually just write this function as $C$, making the result $\operatorname{sgn}x\operatorname{arcosh}|x|+C$.

We get the same subtlety with $\int\frac1xdx$ (see also here). (In both cases, the complex equivalent of the problem changes the details somewhat.) Clearly, to be able to just get $\operatorname{arcosh}x+C$ we need an $x\ge1$ restriction.

Answer 4

We can integrate the given integral $$\int \frac{1}{\sqrt{x^{2}-1}} \,\mathrm dx,$$ on only some subset of either $[1,\infty)$ or $(-\infty,1]$ but not both.

Noting that $\cosh$'s value is at least $1,$ the substitutions $x=\cosh (y)$ and $x=-\cosh (y)$ work for the former and latter, respectively.

Noting that $\cosh y$ is a many-to-one function, $[0,\infty)$ is conventionally adopted as its principal domain. Thus, it is always the case that $$\sqrt{\sinh^2 (y)}=|\sinh (y)|=\sinh (y).$$

Answer 5

Rather than making the substitution directly, you can note that $$\int\frac{\mathrm{d}x}{\sqrt{x^2-1}}=\int\frac{\mathrm{d}x}{\sqrt{\cosh[\cosh^{-1}(x)]^2-1}}=\int\frac{\mathrm{d}x}{\sqrt{\sinh[\cosh^{-1}(x)]^2}}=\int\frac{\mathrm{d}x}{|\sinh[\cosh^{-1}(x)]|}.$$ Now, since $\cosh^{-1}(x)\geq0$ everywhere, it follows that $|\sinh[\cosh^{-1}(x)]|=\sinh[\cosh^{-1}(x)],$ hence $$\int\frac{\mathrm{d}x}{|\sinh[\cosh^{-1}(x)]|}=\int\frac{\mathrm{d}x}{\sinh[\cosh^{-1}(x)]}.$$ To actually evaluate the expression, note that $\mathrm{d}[\cosh^{-1}(x)]=\frac{\mathrm{d}x}{\sqrt{x^2-1}},$ so $$\int\frac{\mathrm{d}x}{\sinh[\cosh^{-1}(x)]}=\int\frac{\sqrt{x^2-1}\cdot\mathrm{d}x}{\sinh[\cosh^{-1}(x)]\cdot\sqrt{x^2-1}}=\int\frac{\sqrt{x^2-1}\cdot\mathrm{d}[\cosh^{-1}(x)]}{\sinh[\cosh^{-1}(x)]}=\int\frac{\sinh[\cosh^{-1}(x)]\cdot\mathrm{d}[\cosh^{-1}(x)]}{\sinh[\cosh^{-1}(x)]}=\int\mathrm{d}[\cosh^{-1}(x)].$$ At this point, it would be pointless to do the substitution $y=\cosh^{-1}(x),$ but this method of symbolic manipulation highlights what is going on behind the scenes whenever you perform a substitution. You are not really doing the subsitution $x=\cosh(y),$ you are really doing the substitution $y=\cosh^{-1}(x),$ which happens to imply the former, though the former does not imply the latter. It is because of this that you can ignore the absolute value: $y\geq0$ necessarily under that substitution.

Everything I just said is applicable for $x\gt1.$ When taking the antiderivative for $x\lt-1,$ it becomes a bit different, because then, $x=\cosh[\cosh^{-1}(x)]$ is false. Instead, note that $\sqrt{x^2-1}=\sqrt{(-x)^2-1}=\sqrt{\cosh[\cosh^{-1}(-x)]^2-1}=\sqrt{\sinh[\cosh^{-1}(-x)]^2}=|\sinh[\cosh^{-1}(-x)]|=\sinh[\cosh^{-1}(-x)].$ Notice, though, that now, you do the substitution $y=\cosh^{-1}(-x)$ instead. Notice that $\mathrm{d}[\cosh^{-1}(-x)]=-\frac1{\sqrt{x^2-1}},$ so $$\frac{\mathrm{d}x}{\sinh[\cosh^{-1}(-x)]}=\frac{-\sqrt{x^2-1}\cdot\mathrm{d}x}{\sinh[\cosh^{-1}(-x)]\cdot-\sqrt{x^2-1}}=-\frac{\sinh[\cosh^{-1}(-x)]\cdot\mathrm{d}[\cosh^{-1}(-x)]}{\sinh[\cosh^{-1}(-x)]}=-\mathrm{d}[\cosh^{-1}(-x)].$$ The absolute value still does not matter, since it is still the case that $\cosh^{-1}(-x)\geq0,$ but now, you do have to take care of the negative sign for $x\lt-1.$ The antiderivatives of $\frac1{\sqrt{x^2-1}}$ are not $\cosh^{-1}(x)+C,$ but the piecewise $\cosh^{-1}(x)+A$ for every $x\gt1$ and $-\cosh^{-1}(-x)+B$ for every $x\lt-1.$