Find integral of sqrt((1-x)/(1+x))

[ae4jm]

[SOLVED] Find integral of sqrt((1-x)/(1+x))

Homework Statement

$$\int\sqrt{\frac{1-x}{1+x}}dx$$

Homework Equations

The Attempt at a Solution

I have started by multiplying by $$\sqrt{1-x}$$ in the numerator and denominator. Then I separated the two fractions to get
$$\int\frac{1}{1-x^2}dx$$$$-\int\frac{x}{1-x^2}dx$$ I'm stuck here! Any help is greatly appreciated!

 

[nicksauce]

Well the first one can be solved by partial fractions (among other methods) and the second one can be solved by a simple substitution.

 

[arunbg]

Please check your multiplication for the denominator. Although wrong, the integrals you have are also easily integrable, first term with trigonometric substitution and the second term with ordinary variable substitution.

 

[ae4jm]

The denominator should be sqrt(1-x^2) instead of just 1-x^2, correct? My easy algebra mistake!

 

[nicksauce]

Right... you can still solve with a trig sub, and a regular sub.

 

[ae4jm]

Thanks to all, does this look correct?

This is what I got:

$$=sin^-^1x+2\sqrt{1-x^2}+c$$

Thanks to everyone!

 

[D H]

You are almost there. It's always a good idea to double check your integration by computing the derivative your result. If you do so, you will see you have a slight mistake.

 

[ae4jm]

I believe my 2 should have canceled out with the $$\frac{1}{2}$$

I'm now getting the answer $$=sin^-^1x+\sqrt{1-x^2}+c$$

After following your help, I did get the integral that I started with after I split it into two fractions. Thanks for your help and for catching my mistake!

 

[D H]

Ahh. Much better.

EDIT:
Now that you know what the answer is supposed to be, you should go back to your derivation to see where you dropped a factor of 1/2 (or added a factor of 2).