# How Do I integrate 1/Sqrt[ x^2-1]

• Mar 22nd 2006, 04:50 AM
guess
How Do I integrate 1/Sqrt[ x^2-1]

integrate 1/Sqrt[ x^2-1]
• Mar 22nd 2006, 05:07 AM
CaptainBlack
Quote:

Originally Posted by guess

integrate 1/Sqrt[ x^2-1]

Try the change of variable:

$\displaystyle \cosh(u)=x$

RonL
• Mar 22nd 2006, 05:21 AM
guess
if there other method??
• Mar 22nd 2006, 05:29 AM
CaptainBlack
Quote:

Originally Posted by guess
if there other method??

Why would you want another method, this change of variable makes the
problem trivial.

All you need to know is that:

$\displaystyle (\cosh(x))^2-(\sinh(x))^2=1$

and that:

$\displaystyle \frac{d}{dx} \cosh(x)=\sinh(x)$

RonL
• Mar 22nd 2006, 05:35 AM
guess
it's because i haven got to learn about cosh...
• Mar 22nd 2006, 07:28 AM
CaptainBlack
Quote:

Originally Posted by guess
it's because i haven got to learn about cosh...

My good friend Mephistopheles suggests that I recommend that you try the
substitution:

$\displaystyle \frac{e^u+e^{-u}}{2}=x$

:cool:

RonL
• Mar 22nd 2006, 07:37 AM
TD!
If you don't want to use hyperbolic functions because you don't know them, an alternative is using the identity $\displaystyle \sec ^2 y = 1 + \tan ^2 y$ to substitute $\displaystyle x = \sec y$.

It won't be as easy as with the hyperbolic functions, but I assume you do know these basic trigonometric functions.
• Mar 24th 2006, 06:24 AM
guess
Thanks Bro,

I solved it using the trigo function substitution