The Integral Trigonometric substitution better explanation

Question

The integral of $$ \int \frac{1}{\sqrt{1-x^{2}}} dx $$ with substitution $x=\sin(\theta)$ is $$ \int \frac{\cos(\theta)}{\sqrt{\cos^{2}(\theta)}} d \theta$$ If I notice, this can take two form, since it can be written as $$ \int \frac{\cos(\theta)}{|\cos(\theta)|} d \theta$$ But why some textbooks (from what I have seen, for example : Schaum's outline of Calculus) does not account this? which means only take the positive value $$ \int \frac{\cos(\theta)}{\sqrt{\cos^{2}(\theta)}} d \theta = \int d \theta$$ Or, am i missing something here?

I presume that it is for simplicity in explanation. If so, which is the better one to be explained to freshmen students? Thanks.

Answer 1

The reason is very simple: a substitution must be defined with a bijective function. $\sin \theta$ is not bijective, so you have to restrict it to an interval on which it is bijective. The usual choice is the interval $\bigl[-\frac\pi2,\frac\pi2\bigr]$ – and on this interval, it happens the cosine is non-negative.

Other than that, evey well-bred person should know the derivative of $\arcsin x\;$ is $\;\dfrac1{\sqrt{1-x^2}}$, so the substitution is rather trivial.

Answer 2

Note that we must have $-1 < x < 1$. If we use $x = \sin \theta$, then we can assume that $-\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}$. It just so happens that $\cos \theta > 0$ in that region.