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Integral with square root

  1. Jun 7, 2005 #1
    Hi,

    I'm trying to find this integral:

    [tex]
    \int \frac{x}{(x+1)\sqrt{1-x^2}}\ dx
    [/tex]

    Because [itex]1-x^2[/itex] has two different real solutions, I can write

    [tex]
    \sqrt{ax^2 + bx + c} = \sqrt{-a}(x_2 - x)\sqrt{\frac{x-x_1}{x_2 - x}}
    [/tex]
    so
    [tex]
    \sqrt{1-x^2} = (-1 - x)\sqrt{\frac{x - 1}{-1 - x}} = (-1 - x)\sqrt{\frac{1-x}{1+x}}
    [/tex]

    I used this substitution:
    [tex]
    t = \sqrt{\frac{1-x}{1+x}}
    [/tex]

    It gives
    [tex]
    x = \frac{1-t^2}{1 + t^2}
    [/tex]

    [tex]
    x + 1 = \frac{2}{1+t^2}
    [/tex]

    [tex]
    dx = \frac{-4t}{(1+t^2)^2}
    [/tex]

    So
    [tex]
    \int \frac{x}{(x+1)\sqrt{1-x^2}}\ dx = \int \frac{ \frac{1-t^2}{1+t^2}}{\left(\frac{2}{1+t^2}\right)\left(\frac{-2}{1+t^2}\right)t}\ \ \frac{-4t}{(1+t^2)^2}\ dt
    [/tex]
    [tex]
    = \int \frac{1-t^2}{1+t^2}\ dt = \int \frac{1}{1+t^2}\ dt - \int \frac{t^2}{1+t^2}\ dt = \arctan t -\ \int \frac{t^2}{1+t^2}\ dt
    [/tex]

    Damn I know I should be able to solve this integral, but I don't know how, maybe it's too late for me... :frown:

    Btw the correct result should be:

    [tex]
    \int \frac{x}{(x+1)\sqrt{1-x^2}}\ dx = \sqrt{\frac{1-x}{1+x}} + 2\arctan \sqrt{\frac{1+x}{1-x}} + C
    [/tex]

    Thank you.
     
    Last edited: Jun 7, 2005
  2. jcsd
  3. Jun 7, 2005 #2

    GCT

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    Why don't you try a simple trig substitution and go from there? You can first get rid of the square root and try to simplify.

    [tex]
    \int \frac{x}{(x+1)\sqrt{1-x^2}}\ dx
    [/tex]

    [tex] \int \frac{sin@cos@~d@}{(sin@cos@+cos@)} [/tex]

    Next try multiplying by [tex] \frac{(cos@-sin@cos@)}{cos@-sin@cos@} [/tex]
     
  4. Jun 7, 2005 #3
    Thank you GCT, it's interesting idea, but it seems it leads to [itex]t = \tan \frac{x}{2}[/itex] substitution, which is little tricky in my experience...could you look please at my original solution and tell me what's wrong there?

    Thanks!
     
  5. Jun 7, 2005 #4

    dextercioby

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    A quick glance

    [tex] x+1=\frac{2}{1+t^{2}} [/tex]

    ,assuming "x" was right in the first place.

    Daniel.
     
  6. Jun 7, 2005 #5
    Thank you Daniel! How stupid I am..So I'm gonna go solve it again, hopefully it will be ok now :smile:
     
  7. Jun 7, 2005 #6
    I edited my initial post now, could you now help me how to solve that?
     
  8. Jun 7, 2005 #7

    dextercioby

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    [tex] \int \frac{t^{2}}{1+t^{2}} \ dt=\int \frac{1+t^{2}-1}{1+t^{2}} \ dt [/tex]

    Daniel.
     
  9. Jun 7, 2005 #8
    Thank you I noticed it already :) Now the result seems ok, except for sign and similar technicalities..
     
  10. Jun 7, 2005 #9

    dextercioby

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    You can check the sign out.It looks okay to me.


    Daniel.
     
  11. Jun 7, 2005 #10
    Well, the right result should be
    [tex]
    \int \frac{x}{(x+1)\sqrt{1-x^2}}\ dx = \sqrt{\frac{1-x}{1+x}} + 2\arctan \sqrt{\frac{1+x}{1-x}} + C
    [/tex]

    but I have
    [tex]
    2 \arctan \sqrt{\frac{1-x}{1+x}} - \sqrt{\frac{1-x}{1+x}}
    [/tex]
     
  12. Jun 7, 2005 #11

    dextercioby

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    I don't have time now,i'll take a deeper look later.

    Daniel.
     
  13. Jun 7, 2005 #12
    [tex]\int \frac{x}{(x+1)\sqrt{1-x^2}}\ dx=\int \frac{x+1-1}{(x+1)\sqrt{1-x^2}}\ dx=\int \frac{1}{\sqrt{1-x^2}}\ dx-\int \frac{1}{(x+1)\sqrt{1-x^2}}\ dx[/tex]
    First one is obvious. For second [tex]t=\frac{1}{x+1}[/tex] will lead to simple solution.
    The answer then is different from what you've given, but Mathematica says that it's absolutely correct.
     
  14. Jun 7, 2005 #13

    GCT

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    Twoflower, in case you're interested

    [tex] \int \frac{sin@cos@~d@}{(sin@cos@+cos@)} * \frac{(cos@-sin@cos@)}{cos@-sin@cos@} [/tex]

    [tex] \int \frac{sin@cos^{2}@-sin^{2}cos^{2}@~d@}{cos^{2}@-sin^{2}@cos^{2}@} [/tex]

    [tex] \int \frac{sin@cos^{2}@}{cos^{2}@-sin^{2}@cos^{2}@}~-~ \int \frac{sin^{2}@cos^{2}@~d@}{cos^{2}@-sin^{2}@cos^{2}@}[/tex]

    [tex] \int \frac{sin@~d@}{cos^{2}@}~-~ \int \frac{sin^{2}@~d@}{cos^{2}@} [/tex]

    The left term can be resolved using substitution. The right term simplifies to
    [tex] \int \frac{1}{cos^{2}@}~-~\int d@ [/tex]

    anyone see any mistakes? Please point them out.
     
  15. Jun 8, 2005 #14
    [tex]\int \frac{sin@~d@}{cos^{2}@} = -\int \frac{d(\cos@)}{cos^{2}@}[/tex]

    It's all right with your solution GCT. At least it's not so artificiall as which Twoflower made.
     
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