# Why we don't use absolute value bars in the trigonometric substitution of indefinite integral?

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 $$\sqrt{u^2}=|u|$$ ?

• $\sqrt{u^2} = |u|$ is false for complex numbers. Except in the most elementary courses, when doing analysis you will want to consider complex numbers. Apr 12, 2021 at 10:28

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.
• 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? Apr 12, 2021 at 4:32