<s>[INST] 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?
 [/INST] Your book is being very careful about the fact that $\sqrt{1-\sin^2\theta} = \cos\theta$ is only true when $\cos \theta \geq 0$. Of course, in this problem it doesn't matter, since $-\frac{\pi}{2} < \theta \leq \frac{\pi}{2}$.
Note that you could just as well have chosen the substitution $x = -3\sin \theta$. It is a good exercise to check that this gives you the same answer as $x=3\sin \theta$.
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