Integrating 1/(1-x^2)^(1/2) using Substitution and Understanding Constant C

[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?

 

[vela]

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 $\tanh^2 u$.

 

[phospho]

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]

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]

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.

 

[phospho]

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!