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For example let's say we have the following indefinite integral $$I=\int\sqrt{1-x^2}dx$$We evaluate it by using trigonometric substitution $x=\sin\theta$:

$$I=\int\sqrt{1-\sin^2\theta}\cos\theta d\theta$$ Here we use $\sqrt{1-\sin^2\theta}=\cos\theta$ rather than $\sqrt{1-\sin^2\theta}=|\cos\theta|$. but why? isn't $\sqrt{u^2}=|u|$ ?

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  • $\begingroup$ $\sqrt{u^2} = |u|$ is false for complex numbers. Except in the most elementary courses, when doing analysis you will want to consider complex numbers. $\endgroup$ – GEdgar Apr 12 at 10:28
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As a function to $\mathbb{R}$, the domain is $[-1, 1]$. So, if $x= \sin(\theta)$, then $\theta \in [-\pi/2, \pi/2]$, and $\cos(\theta )$ is always positive on that interval.

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  • $\begingroup$ Oh I got it. that's because by taking $x=\sin\theta$ we have $\theta=\sin^{-1}x$ and it accepts $\theta\in[-\frac{\pi}2,\frac{\pi}2]$. Am I right? $\endgroup$ – Amirali Apr 12 at 4:32
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    $\begingroup$ @amirali Yes, that's a good way to think of it. $\endgroup$ – David Lui Apr 12 at 4:49

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