Trig substitution; why can we ignore the absolute value?

This is a good question. It is a very easy mistake to assume that $\sqrt{u^2} = u$ (whether the $u$ in question is $\cos^2 \theta$ or something else). Of course the equation is true if $u$ is positive, but then we must have a good reason to say $u$ is positive.

So we started with the integral $$\int \frac{x}{\sqrt{1-x^2}}\, dx,$$ and we decide to try the trig substitution $x = \sin\theta$. This gives us $$ \int \frac{\sin \theta}{\sqrt{\cos^2\theta}}\, \cos\theta \, d\theta. $$ But before we go any further, let's consider what kind of values $x$ might have and how our substitution might relate values of $x$ with values of $\theta$ in this new integral.

I'm assuming you are not doing complex analysis here, that is, only real numbers are allowed, not complex numbers. So $\sqrt{1-x^2}$ is defined only for $-1 \leq x \leq 1,$ because if $|x| > 1$ then $1 - x^2$ is negative.

Now if we allow $x$ to vary within the range $-1 \leq x \leq 1,$ what is the correct range of $\theta$? We want a one-to-one relationship between the values of $x$ in $-1 \leq x \leq 1$ and our chosen range of values of $\theta$. For example, $x = -1$ should correspond to exactly one value of $\theta$ within the range of values we allow for $\theta$.

Clearly if $x = -1$ and $\sin\theta = x$, we must have $\theta = -\frac\pi2 + 2n\pi$ where $n$ is an integer. Suppose we choose $n = 0$, so $\theta = -\frac\pi2$. If we increase $\theta$ toward $\frac\pi2$, then $x$ increases toward $1$. So let's let $\theta$ vary within the range $-\frac\pi2 \leq \theta \leq \frac\pi2.$

But if $-\frac\pi2 \leq \theta \leq \frac\pi2$, then $\cos\theta \geq 0$, and the statement that $\sqrt{\cos^2\theta} = \cos\theta$ is justified. Everything proceeds smoothly from there in the way you have already (presumably) seen in your textbook and lectures.

How can we justify the assumption that $-\frac\pi2 \leq \theta \leq \frac\pi2$? It isn't so much an assumption as a definition on our part. When we substitute one variable for another variable, we get to say what range of values of the new variable we want to relate to the range of values the old variable might have, provided that these are consistent with the function we chose to use to relate the two variables.

Now just out of curiosity, suppose we had done the substitution just a little bit differently? We could have tried letting $\theta$ vary within the interval $\frac\pi2 \leq \theta \leq \frac{3\pi}2$. These values of $\theta$ also give us all the values of $x$ we might need. For these values of $\theta$, $\cos \theta \leq 0$, so $\sqrt{\cos^2\theta} = -\cos\theta$. Our integral then becomes $$\int \frac{\sin \theta}{\sqrt{\cos^2\theta}}\, \cos\theta \, d\theta = \int -\sin \theta \, d\theta = \cos\theta + C.$$ Now we want to reverse the substitution in order to get the answer to our original integral in terms of $x$. Since $\cos\theta \leq 0$, the only substitution consistent with our original substitution is $$\cos\theta = -\sqrt{1-x^2}.$$ We conclude that $$\int \frac{x}{\sqrt{1-x^2}}\, dx = -\sqrt{1-x^2} + C,$$ which is the same result we got when we let $-\frac\pi2 \leq \theta \leq \frac\pi2.$

We would also get the same result if we tried $-\frac{3\pi}2 \leq \theta \leq -\frac\pi2$ or $\frac{7\pi}2 \leq \theta \leq \frac{9\pi}2$. That's good; the solution to the integral is not a mere accident of exactly which substitution we chose.

We are letting $\theta=\arcsin x$. So $\theta$ ranges from $-\pi/2$ to $\pi/2$. Over this interval the cosine is non-negative.