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Interesting integral

  1. Mar 29, 2006 #1
    Dear community

    I'm trying to get a grip on this integral:
    [tex]\int \frac{\sqrt{1-x}}{\sqrt{x}-1} dx[/tex].
    I tried substituting [tex]x=\sin^{2}(u)[/tex], which leaves me (standing) with
    [tex]\int \frac{\sin(u)\cos^{2}(u)}{\sin(u)-1} du[/tex].

    But I just can't solve it, no matter which way I try.
    I would be thankful for every kind of hint/explanation.
    Best regards...Cliowa
     
  2. jcsd
  3. Mar 29, 2006 #2

    arildno

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    Well,.when nothing else works, make the s=tan(u/2) substitution:
    [tex]\cos(u)=\cos^{2}\frac{u}{2}-\sin^{2}\frac{u}{2}=\cos^{2}(\frac{u}{2})(1-s^{2})=\frac{(1-s^{2})}{\sec^{2}\frac{u}{2}}=\frac{(1-s^{2})}{1+s^{2}}[/tex]
    [tex]\frac{\sin(u)}{\sin(u)-1}=\frac{2\sin\frac{u}{2}\cos\frac{u}{2}}{2\sin\frac{u}{2}\cos\frac{u}{2}-\cos^{2}\frac{u}{2}-\sin^{2}\frac{u}{2}}=\frac{2s}{2s-s^{2}-1}=-\frac{2s}{(s-1)^{2}}[/tex]

    [tex]u=2\arctan(s)\to{du}=\frac{2ds}{1+s^{2}}[/tex]

    Collecting, we get to evaluate the integral:
    [tex]-\int\frac{4s(1+s)^{2}}{(1+s^{2})^{3}}ds[/tex]
     
    Last edited: Mar 29, 2006
  4. Mar 29, 2006 #3

    TD

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    Before introducing trigonometry, I'd substitute [tex]y = \sqrt{1-x}[/tex].
     
  5. Mar 29, 2006 #4

    VietDao29

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    You are forgetting a factor of 2. It should read:
    [tex]2 \int \frac{\sin(u)\cos^{2}(u)}{\sin(u)-1} du[/tex] instead.
    Now by using the Pythagorean Theorem, we have:
    cos2x = 1 - sin2x = -(sin x - 1) (sin x + 1)
    So you'll have:
    [tex]2 \int \frac{\sin(u)\cos^{2}(u)}{\sin(u)-1} du = -2 \int \sin u (\sin u + 1) du[/tex].
    You can go from here, right? :)
     
  6. Mar 29, 2006 #5

    arildno

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    That was a bit simpler than mine, VietDao..:frown:
     
  7. Mar 30, 2006 #6
    Thank you very much, arildno, VietDao29 for your great help.
    I figured out the rest on my own.
    And, although s=tan(u/2) is a bit more complicated I think it's always good to have to ways to go.
    Best regards...Cliowa
     
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