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

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?

## Answers and Replies

vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
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*.
Yes, you're right. It's the constant. With the correct choices, those two expressions are equal.

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?
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##.

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##.

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.

vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
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.

vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
now I can integrate sech(u) by using the exponential form, however this becomes messy and I end up getting something more complex.
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.

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.

Thanks for your help!