# How to solve $\int \frac{1}{\sqrt{\frac{C}{x^2}-1}}\;dx\;\;$

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?

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

## 1 Answer

$$\int \frac{1}{\sqrt{\frac{C}{x^2}-1}}dx = \int \frac{1}{\sqrt{C-x^2}}\frac{1}{\frac{1}{x}}dx=\\ \int \frac{xdx}{\sqrt{C-x^2}}$$ us the sub $C-x^2 = u$

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