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    Find integral of
    √[(1-x)/(1+x)] dx
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    (Original post by Ano123)
    Find integral of
    √[(1-x)/(1+x)] dx
    my first attempt would be 2/(1+x) = cosh2u
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    (Original post by Ano123)
    Find integral of
    √[(1-x)/(1+x)] dx
    Try x=\cos u substitution and then use trig identities.
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    How about u = 1-x ?
    That will turn your integrand into u/u-2 which I assume can be simplified by long division?
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    (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..
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    (Original post by notnek)
    But you still have the square root..
    That is the tiniest square root I have never seen in my life..
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    (Original post by oShahpo)
    That is the tiniest square root I have never seen in my life..
    They're the deadliest kind.
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    Try x=cos2θ
    Simplifies very nicely then.
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    Even better, try multiplying up and down by root(1-x), you'll get two fractions which you can then integrate by substitution.
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    (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.
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    (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
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    (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.
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    (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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