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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?

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    $\begingroup$ $|3|$ is the same value as 3, so the modulus signs are unnecessary. Also - you can't "choose" 3 to be positive or negative - it is positive $\endgroup$
    – Old John
    Jan 1, 2013 at 22:09
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    $\begingroup$ Something else to take note of, the range that you chose: $-\frac {\pi}{2} \leq \theta \leq {\pi}{2}$. Had you decided to choose a range where $\cos \theta$ is negative, then you would have to use $ - 3 \cos \theta$, which will now be a positive value. $\endgroup$
    – Calvin Lin
    Jan 1, 2013 at 22:17

3 Answers 3

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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: $\,x\, =\,\pm \sqrt{a}\,$.

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The core of your problem seems to be a confusion about the meaning of $\left| \cdot \right|$. $\left| x \right|$ is not the same as $\pm x$. Instead:

  • $\left| x \right| = -x$ if $x \le 0$
  • $\left| x \right| = x$ if $x \ge 0$

Now, because $\sqrt{\cdot}$ is a real-valued function, it can only return one of the quadratic roots of its argument. We've chosen that it's the nonnegative one, so $\sqrt{x^2}=\left|x\right|$, while the roots of $x^2-a$ are $\pm \sqrt{a}$.

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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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