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 $\sqrt{u^2}=|u|$ ?
Best Answer
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.