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# Integration- by sub?? watch

1. Find integral of
√[(1-x)/(1+x)] dx
2. (Original post by Ano123)
Find integral of
√[(1-x)/(1+x)] dx
my first attempt would be 2/(1+x) = cosh2u
3. (Original post by Ano123)
Find integral of
√[(1-x)/(1+x)] dx
Try substitution and then use trig identities.
4. How about u = 1-x ?
That will turn your integrand into u/u-2 which I assume can be simplified by long division?
5. (Original post by oShahpo)
How about u = 1-x ?
That will turn your integrand into u/u-2 which I assume can be simplified by long division?
But you still have the square root..
6. (Original post by notnek)
But you still have the square root..
That is the tiniest square root I have never seen in my life..
7. (Original post by oShahpo)
That is the tiniest square root I have never seen in my life..
8. Try x=cos2θ
Simplifies very nicely then.
9. Even better, try multiplying up and down by root(1-x), you'll get two fractions which you can then integrate by substitution.
10. (Original post by oShahpo)
Even better, try multiplying up and down by root(1-x), you'll get two fractions which you can then integrate by substitution.
Can you show me what that leads to, I'm not sure that will work.
11. (Original post by Ano123)
Find integral of
√[(1-x)/(1+x)] dx
(Original post by TeeEm)
my first attempt would be 2/(1+x) = cosh2u
actually the quickest way to do this is to multiply by (1-x), top and bottom under the root
simplify
split the fraction
and you get 2 very easy antiderivatives
12. (Original post by Ano123)
Can you show me what that leads to, I'm not sure that will work.
It does, you get two fractions. One can be done by a very simple substitution and the other can be done by knowledge of integration of Inverse Hyperbolics.
13. (Original post by oShahpo)
It does, you get two fractions. One can be done by a very simple substitution and the other can be done by knowledge of integration of Inverse Hyperbolics.
Indeed after trying it, it does appear to be the easiest way, although using a trig sub can work as well.

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