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Very hard integral. please help me guys

  1. Apr 17, 2013 #1
    hi guys!
    can you please help me with this integral? Ive tried some subs but i realy dont know how to start....
    Thank you !!!!
     

    Attached Files:

  2. jcsd
  3. Apr 17, 2013 #2

    mfb

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    Expanding with ##\sqrt{\ln(9-x)}-\sqrt{\ln(3+x)}## gives
    $$\frac {\sqrt{\ln(9-x)}\left(\sqrt{\ln(9-x)}-\sqrt{\ln(3+x)}\right)} {\ln(9-x)-\ln(3+x)}$$

    That can be split into two parts, where one part looks easy and the other part can be simplified significantly. I don't know if that will lead to a solution, however.
     
  4. Apr 17, 2013 #3
    first of all thank you!
    Ive tried this and it lead me nowhere..
     
  5. Apr 17, 2013 #4

    mfb

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    Hmm okay, I don't see how to solve the remaining part either.

    An interesting observation: 3+x goes from 5 to 7, while 9-x goes from 7 to 5. There could be some symmetry to exploit, but I did not find it yet.
     
  6. Apr 18, 2013 #5
    Thats actually very interesting because according to the answers the answer is 1.
    Can someone help me...?
     
  7. Apr 18, 2013 #6

    Curious3141

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    Hint: substitute ##\displaystyle x = 6-y##.

    EDIT: numerous glitches in the PF software are making it difficult to post. Basically, I wanted to add: don't try to simplify the integrand, etc. Just do the sub, and see what happens to the bounds. This is actually a very simple problem in disguise.
     
    Last edited: Apr 18, 2013
  8. Apr 18, 2013 #7
    I cant remove the disguise... :(
     
  9. Apr 18, 2013 #8
    this is what i get
     

    Attached Files:

  10. Apr 18, 2013 #9

    mfb

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    Ah nice, that is the right way to solve it.
    @omri_mar: You can modify this new integrand to look similar to the old one now.
     
  11. Apr 18, 2013 #10
    I dont get it. How can I continue?
     
  12. Apr 18, 2013 #11
    Try adding together the new integral and the old one
     
  13. Apr 18, 2013 #12
    what do you mean?
     
  14. Apr 18, 2013 #13

    mfb

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    You have an equation (old integral)=(new integral). Try to write (new integral) as (something)-(old integral). This gives you an equation like 2*(old integral)=(something), and (something) is easy to calculate.

    Don't write PMs, please.
     
  15. Apr 19, 2013 #14

    Curious3141

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    Sorry for the late reply - I just noticed your bounds are off in the attachment to post 8. Your bounds should be reversed (identical to the original integral). Don't forget that ##dy = -dx##. Other than that, the other posters have already told you how to proceed. Remember that in a definite integral, the variable of integration is basically just a dummy variable, so whether it's x or y, it doesn't matter. So just replace the y in the new integral with x, and add it to the original integral. Now you have twice the original integral, and it's equal to something very simple.
     
  16. Apr 19, 2013 #15

    SammyS

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    (Most of post deleted in Edit. It was pretty much redundant with Curious3141 post. )

    Now modify the numerator by adding and subtracting ... (something clever) .
     
    Last edited: Apr 19, 2013
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