Why do we choose $3$ to be positive after $\sqrt{9 - x^2}$ in the following substitution? The integral $$\int \frac{\sqrt{9 - x^2}}{x^2}dx$$ is solved in my book by letting $x = 3\sin\theta$ where $-\frac {\pi}{2} \le \theta \le \frac {\pi}{2}$. 
Then, $dx = 3\cos\theta\,d\theta$ 
and, $$\sqrt{9-x^2} = 3|\cos\theta| = 3\cos\theta$$
So, $$\int \frac{\sqrt{9 - x^2}}{x^2}dx = \int \cot^2 \theta \ d\theta = -\cot\theta - \theta + C$$
Returning to the original variable, $$\int \frac{\sqrt{9 - x^2}}{x^2}dx = -\frac {\sqrt{9 - x^2}}{x} - \sin^{-1}\left(\frac{x}{3}\right) + C$$
I don't understand why $\sqrt{9-x^2} = 3|\cos\theta| = 3\cos\theta \,$ instead of $\sqrt{9-x^2} = |3||\cos\theta| = |3|\cos\theta$. I feel like I have problems understanding this because I am not sure what is the purpose of the absolute value signs in this case, are they to indicate that, for example, $|\cos\theta| = \pm\cos\theta$? If that's the case, why do we choose $3$ to be positive instead of negative?

Think about it: How would using $\;|\,3\,|\;$ change the result?$\quad$ After all,  $\;|\,3\,| \;= \quad?$
Also note: $\,|\cos\theta| \,= \,\cos\theta \,\ge \,0\;$ since the range of $\,\theta\,$ is $\;\;-\dfrac{\pi}{2} \,\le\, \theta \,\le \dfrac{\pi}{2}$.

You seem to be confused about the what "$|\;\cdot\;|$" means. 
$\;| \,a \,|$ is not the same as $\,\pm a\,$. 
Rather, we define


*

*$|\, a \,| = -a\,$ if $\,a \lt 0$.

*$|\, a\, | = \;\,\;a\,$ if $\,a \geq 0$.
This ALWAYS returns a non-negative result.
By definition, $\,\sqrt{\;\;}\,\;$ returns only the non-negative root of a square or quadratic. So $\,\sqrt{a^2}=|a|,\,$ while solving for, say, $\,x^2-a = 0\,$ we obtain two roots:  $\,x\, =\,\pm \sqrt{a}\,$.
