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

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I'm trying to integrate [itex] \dfrac{1}{\sqrt{1-x^2}} [/itex] 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 [itex] \displaystyle\int \dfrac{sech^2u}{1-sech^2u}\ du [/itex], where can I go from here?
 

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vela
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I'm trying to integrate [itex] \dfrac{1}{\sqrt{1-x^2}} [/itex] 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 [itex] \displaystyle\int \dfrac{sech^2u}{1-sech^2u}\ du [/itex], 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##.
 
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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:

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

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