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How does one solve the following integral:

$$\int \frac{1}{\sqrt{\frac{C}{x^2}-1}}\;dx\;\;,$$

where $C$ is some constant. Should substitution be used here?

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    $\begingroup$ A difference of squares may suggest a trigonometric substitution. $\endgroup$ – GEdgar Jan 12 '15 at 14:26
  • $\begingroup$ Euler substitutions $\endgroup$ – Pp.. Jan 12 '15 at 14:26
  • $\begingroup$ $t=x^2$ is sufficient. $\endgroup$ – mickep Jan 12 '15 at 14:26
  • $\begingroup$ When math education will stop teaching the "trigonometric substitutions"? How many generations of students damaged! $\endgroup$ – Pp.. Jan 12 '15 at 14:28
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    $\begingroup$ If $C \leq 0$ the radical is negative. If $C > 0$, the domain of integrability is $-\sqrt{C} <x< \sqrt{C}$. $\endgroup$ – Alex Silva Jan 12 '15 at 15:39
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$$ \int \frac{1}{\sqrt{\frac{C}{x^2}-1}}dx = \int \frac{1}{\sqrt{C-x^2}}\frac{1}{\frac{1}{x}}dx=\\ \int \frac{xdx}{\sqrt{C-x^2}} $$ us the sub $C-x^2 = u$

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    $\begingroup$ Should be $|x|$ on top. $\endgroup$ – KittyL Jan 12 '15 at 14:30
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    $\begingroup$ I don't claim this is wrong, but just that one should be a bit careful. $\sqrt{x^2}=|x|$. $\endgroup$ – mickep Jan 12 '15 at 14:30

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