# Integral of 1/(1-x^2)^(1/2)

1. Mar 7, 2013

### phospho

I'm trying to integrate $\dfrac{1}{\sqrt{1-x^2}}$ with respect to x, using an appropriate substitution

I've tried using:

x = sin(u) which leads me to the integral being arcsin(x) + c
x = cos(u) which leads me to the integral being -arccos(x) + c

however I have drawn arcsin(x) and -arccos(x) and they are not the same, is this because of the constant C? I.e the constant is not necessarily the same in both of the integrals? (The graphs are on *close*.

Also, I've tried using the substitution x = tanh(u), this lead me to $\displaystyle\int \dfrac{sech^2u}{1-sech^2u}\ du$, where can I go from here?

2. Mar 7, 2013

### vela

Staff Emeritus
Yes, you're right. It's the constant. With the correct choices, those two expressions are equal.

You have the identity $\cosh^2 u - \sinh^2 u = 1$. You should be able to show the denominator is equal to $\tanh^2 u$.

3. Mar 7, 2013

### phospho

thanks,

for the substitution of x = tanh(u) I seem to have gone wrong some where and I do not end up with what I originally posted:

$x = tanh(u)$
$\dfrac{dx}{du} = sech^2u$
$\displaystyle\int \dfrac{1}{\sqrt{1-x^2}}\ dx = \displaystyle\int \dfrac{sech^2u}{\sqrt{1 - tanh^2u}}\ du = \displaystyle\int sech(u)\ du$ now I can integrate sech(u) by using the exponential form, however this becomes messy and I end up getting something more complex.

4. Mar 7, 2013

### vela

Staff Emeritus
I should have looked at your work more carefully. I didn't notice the mistake either.

You can get quite different looking expressions that at first glance don't appear to be equal, but you can eventually show that they are.

5. Mar 8, 2013

### vela

Staff Emeritus
I found $\int \text{sech }u\,du = \tan^{-1} \sinh u$ while Mathematica gave me $\int \text{sech }u\,du = 2\tan^{-1} \tanh(u/2)$. If you plug in $u=\tanh^{-1} x$, you get expressions which don't seem to be equal to arcsin x, but in fact they are, which you can see by plotting the functions.

6. Mar 8, 2013