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In general, when we are trying to remove radicals from integrals, we perform a trigonometric substitution (either a circular or hyperbolic trig function), but often this results in a radical of the form $\sqrt{(f(x))^2}$, with $f$ being an arbitrary trigonometric function.

What most texts tend to do is simply take $\sqrt{(f(x))^2} = f(x)$, without the absolute value of |f(x)|, and the texts do not offer any motivation as to why $\sqrt{(f(x))^2} = f(x) \neq |f(x)|$. I would have assumed the correct way to proceed would be $\sqrt{(f(x))^2} = |f(x)|$. Why is this the case?

I'll give an example to show further explain what I'm trying to ask :

$$\text{Integrate} \ \ \ \int{\frac{1}{\sqrt{x^2 + 16}}}\ dx$$

We let $\ x = 4\tan\theta \implies dx = 4\sec^2\theta \ d\theta$ \begin{equation} \label{eq1} \begin{split} \implies\int{\frac{1}{\sqrt{x^2 + 16}}}\ dx & = \int{\frac{4\sec^2\theta \ }{\sqrt{(4\tan\theta)^2 + 4^2}}}\ d\theta \\ & = \int{\frac{4\sec^2\theta \ }{\sqrt{(4^2\sec^2\theta)}}}\ d\theta \\ &= \int{\frac{4\sec^2\theta \ }{4\cdot|\sec\theta \ |}}\ d\theta \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{(*)}\\ \end{split} \end{equation}

What most texts do is omit the absolute value in the last starred step. Thus the denomitor of the integral becomes $\ 4\sec\theta \ $ instead of $4\cdot|\sec\theta \ |$ and there is no need to break the integral up into cases. Why is that so? We have not assumed $\sec\theta > 0$, so how can $|\sec\theta \ | = \sec\theta$?

Perturbative
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1 Answers1

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The substitution you do must be bijective. When you set $x=4\tan\theta$, you're tacitly assuming $\theta\in(-\pi/2,\pi/2)$ and, in this case, $\sec\theta>0$. Thus there's no need of $\lvert\sec\theta\rvert$.

You're right, however, that the possibility of removing the absolute value should be checked.

egreg
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    Of course, it would be equally valid to enforce the same substitution $x=4\tan (\theta)$ for $\theta \in (\pi/2,3\pi/2)$ (or any other branch on which the transformation is one-to-one) for which the secant is always negative. – Mark Viola Jul 29 '16 at 19:22
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    @Dr.MV Yes; the result would not change, of course, once the back substitution is performed. On the other hand, a better statement would be “set $\theta=\arctan(x/4)$”, which would clarify the interval where $\theta$ is supposed to be. – egreg Jul 29 '16 at 20:08
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    Yes, absolutely. I thought that it was useful to add this to the discussion. Oh, and +1. -Mark – Mark Viola Jul 29 '16 at 20:09