Why do we take $\sqrt{f^2(x)} = f(x)$ when integrating by substitution

Question

I've posted a similar question Confusion in finding derivative of $\sqrt{\frac{1-\cos(2x)}{1 + \cos(2x)}}$.

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Consider the following integral: $$\int\sqrt{1 – x^2}\ dx.$$
Putting $x = \sin(\theta):$
$$=\int\sqrt{1 – \sin^2\theta}\ \cos\theta\ d\theta$$
$$=\int\sqrt{\cos^2\theta}\ \cos\theta\ d\theta\tag{1}$$
$$=\int\ \cos\theta\cos\theta\ d\theta\tag{2}$$
$$=\int\cos^2\theta\ d\theta$$
$$…$$
$$…$$
$$…$$

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From step $(1)$ to $(2)$, I don't understand why we take $\sqrt{f^2(x)} = f(x)$ instead of $\sqrt{f^2(x)} =|f(x)|.$ I've seen many such examples like the above, where we ignore the negative values (mainly in substitutions in integration, differentiation, and inverse trigonometric functions).

Answer 1

$$\int\sqrt{1 - x^2}\,\mathrm dx.$$ Putting $x = \sin(\theta):$ \begin{align}&\implies\int\sqrt{\cos^2\theta}\ \cos\theta\,\mathrm d\theta\tag{1}\\ &\implies\int\ \cos\theta\cos\theta\,\mathrm d\theta\tag{2}\\ &\implies\int\cos^2\theta\,\mathrm d\theta\end{align} why we take $\sqrt{f^2(x)} = f(x)$ instead of $\sqrt{f^2(x)} =|f(x)|.$

1. The above is an implicit substitution of the form $x=h(\theta),$ which requires $h$ to be invertible.

   The author has tacitly restricted $\sin\theta$ to its principal domain so that $h$ has domain $$\left[-\frac\pi2,\frac\pi2\right].$$

   Thus, $\cos\theta\ge0;$ so, $\sqrt{\cos^2\theta}=|\cos\theta|$ simply equals $\cos\theta.$

2. A nitpick: the implication symbol $\implies$ connects statements like $x^2+3x=7,$ and is not interchangeable with the symbol $=,$ which connects expressions like $x^2+3x.$

3. As a contrast, this solution opts for the alternative substitution $\displaystyle x=\sin\alpha\quad\left(\alpha\in\left[\frac\pi2,\frac{3\pi}2\right]\right):$ \begin{align}&\int\sqrt{1 - x^2}\, \mathrm dx\\ \\={}&\int\sqrt{\cos^2\alpha}\ \cos\alpha\, \mathrm d\alpha\\ \\= {}&\int\ (-\cos\alpha)\cos\alpha\, \mathrm d\alpha\\ \\= {}&\int(-\cos^2\alpha)\, \mathrm d\alpha.\end{align} It nonetheless gives the same answer (the negative sign appears as this substitution function is decreasing, which flips the integration limits relative to the previous substitution): \begin{align}&\int_{-1}^{1}\sqrt{1 - x^2}\, \mathrm dx \\= {}&\int_{\color{red}{3\pi/2}}^{\color{red}{\pi/2}}(\color{red}-\cos^2\ \color{red}{ \alpha})\, \mathrm d \color{red}{\alpha} \\= {}&\int_{\pi/2}^{3\pi/2}\cos^2\alpha\, \mathrm d\alpha \\= {}&\int_{\pi/2}^{3\pi/2}\cos^2\theta\, \mathrm d\theta.\end{align}