# integrate $\int\sqrt{x^2-1}dx$ unproven step

In order to find $\int\sqrt{x^2-1}dx$ one makes substitutions $x=\sec(\theta)$ , $dx=\sec(\theta)\tan(\theta)d\theta$ and $\sqrt{x^2-1}$ = $\sqrt{\tan^2(\theta)}$. Then you find $\int{\tan^2(\theta) \sec(\theta)}d\theta$ directly as easy as $1+1=2$ it seems from many online sources and even WolframAlpha. But I tried for hours but every way I try, $\sqrt{\tan^2(\theta)}\neq\tan(\theta)$. I already taken into account $\theta$ is a substitution for $\sec^{-1}(x)$, does not make a difference. What do I think wrong? (edit: and also $\sqrt{\tan^2(\theta)} * \tan(\theta)$ != $\tan^2(\theta)$ )

• I suppose that != in your post is suppose to mean "not equal to" you can type this as $\ne$ $\ne$ or $\neq$ $\neq$. – Martin Sleziak Sep 17 '15 at 8:15
• Also there is difference between $\sec^-1(x)$ $\sec^-1(x)$ and $\sec^{-1}(x)$ $\sec^{-1}(x)$ . (You need to do this if subscript/superscript is more than one item.) – Martin Sleziak Sep 17 '15 at 8:17
• For more information about writing math at this site see e.g. here, here, here and here – Martin Sleziak Sep 17 '15 at 8:17
• As Mickep suggests, you can look at the cases $x\ge1$ and $x\le-1$ separately. Some authors avoid this problem by defining the range of the inverse secant function to be $[0,\frac{\pi}{2})\cup[\pi, \frac{3\pi}{2})$, since then $\tan\theta\ge0$ so $\sqrt{\tan^2\theta}=\tan\theta$. – user84413 Sep 17 '15 at 22:18

It's important to note that $\sec(\theta)=\sec(-\theta)=\sec(\theta+\pi)$. So when you perform the substitution $x=\sec(\theta)$, you should specify the range of $\theta$ so that for each $x$, there is only one $\theta$.

For example we could take $\theta\in(0,\pi/2)$ if $x>0$ and $\theta\in (\pi,3\pi/2)$ if $x<0$. And then you can safely imply that $\tan\theta>0$, or $\sqrt{\tan^2\theta}=\tan\theta$.

Indeed, $$\sqrt{\tan^2\theta}\neq \tan \theta$$ in general, but $$\sqrt{\tan^2\theta}=|\tan\theta|= \begin{cases} \tan\theta, & \text{if}\ \tan\theta\geq 0;\\ -\tan\theta, & \text{if}\ \tan\theta<0. \end{cases}$$ since for any real $x$, $$\sqrt{x^2}=|x|$$ This means that you should be a bit careful with the sign.

• Yes, but that means in the rest of the solution you just take for granted you are working with absolute values? And you can do that because the original function is also always positive? Something like this? Because if you look at it step by step I don't see any other direct justification you can just forget about the sign.. – user34909 Sep 17 '15 at 8:22
• It is usually not so nice to keep absolute values, so it is better to think of the variables belonging to some specific domain. Hopefully, one only has to do the calculation once, and the necessary changes are simple. In this particular case, one should remember that the original function is defined for $x>1$ and $x<-1$. Maybe it is best to start working in the domain $x>1$, do all calculations carefully, and then to modify the result to fit $x<-1$ as well. – mickep Sep 17 '15 at 11:11

$\int\sqrt (x^2-1)dx$

(1)Use a substitution of $x=cosh \theta$, $dx=sinh\theta$

=$\int\sqrt(cosh^2-1).(sinh\theta)d\theta$

=$\int\sqrt sinh^2\theta.(sinh\theta)d\theta$

=$\int\sinh^2\theta d\theta$

Using double angle formulae: $cosh 2\theta=1+2sinh^2\theta$

Hence, $sinh^2\theta=\frac{1}{2}(cosh 2\theta-1)$

=$\frac{1}{2}\int (cosh2\theta-1)d\theta$

=$\frac{1}{2}[\frac{sinh 2\theta}{2}-\theta]$

Need to otain final answer in terms of $x$, so from (1), substitute $cosh\theta=x$

in $cosh^2\theta-sinh^2\theta=1$. Hence $sinh\theta=\sqrt(x^2-1)$

Therefore $sinh 2\theta=2cosh\theta sinh\theta=2x\sqrt(x^2-1)$

$\frac{1}{2}[\frac{2x\sqrt(x^2-1)}{2}-cosh^{-1}x]$+C

$\frac{x\sqrt(x^2-1)}{2}-\frac{cosh^{-1}x}{2}$+C

Here is a more complicated, but more rigorous way, of solving this problem.

The function $f(x) = \sqrt{x^2-1}$ has domain $(-\infty,-1]\cup [1,\infty)$. But we will ignore the endpoints and rather think of it as being defined on $(-\infty,-1)\cup (1,\infty)$, because "differentiability" is defined on open intervals. Notice that the domain is disconnected. This means if $F_1(x),F_2(x)$ are anti-derivatives of $f(x)$, then it is no longer true that $F_1(x) = F_2(x) + c$ for some real number $c$.

So instead what we have to do is choose one of these two intervals. The more natural choice is to pick the interval $(1,\infty)$ and treat $f(x)$ as being defined on that interval.

Now let $g(x) = \sec^{-1} x$. This function is defined and differentiable on the same interval $(1,\infty)$. We also know that $g'(x) = \frac{1}{x\sqrt{x^2-1}}$.

Observe that, for $x>1$, $$f(x) = \sqrt{x^2-1} = \frac{x(x^2-1)}{x\sqrt{x^2-1}} = x(x^2-1)g'(x)$$ We also have that, $$(x^2-1) = \sec^2(\sec^{-1} x) - 1 = \sec^2(g(x)) - 1 = \tan^2(g(x))$$ And, $$x = \sec(\sec^{-1} x) = \sec(g(x))$$ Putting it together, $$f(x) = \sec(g(x))\tan^2(g(x)) g'(x)$$

The "substitution rule" therefore tells us that if we find an anti-derivative $F(x)$ of $\sec(x)\tan^2(x)$ then $F(g(x))$ will be anti-derivative of $f(x)$.

As you can see your difficulty does not come up in this more rigorous method.