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Evaluate this integral

  1. Jun 21, 2013 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    [itex] \displaystyle \int^1_0 cot^{-1}(x^2 - x +1)\ dx [/itex]

    2. Relevant equations

    3. The attempt at a solution
    I used this formula

    [itex]2I=\int^b_a f(x)+f(a+b-x)\ dx [/itex]

    But using this method I arrived at the original question. OK, So I tried integrating by parts and it's still useless. Substitution doesn't work either.
     
  2. jcsd
  3. Jun 21, 2013 #2

    tiny-tim

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    hi utkarshakash! :smile:

    have you tried integrating by parts with u = x ?
     
  4. Jun 21, 2013 #3
    [tex]\cot^{-1}(x^2-x+1)=\tan^{-1}\left(\frac{1}{x^2-x+1}\right)=\tan^{-1}\left(\frac{1}{1-x(1-x)}\right)[/tex]

    :smile:
     
  5. Jun 21, 2013 #4

    haruspex

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    In case Pranav-Arora's hint is still obscure, compare it with the expansion of tan(a+b).
     
  6. Jun 22, 2013 #5
    Firstly write it as,

    tan-1(1/(x2-x+1) = tan-1{(x-(x-1))/(1+x(x-1))}

    Don't you see an obvious relation of tan-1A - tan-1B formula from here ? You may proceed..

    It was an easy question though. :smile:

    Method II :

    Let f(x) =cot−1(x2−x+1)

    Write as,

    f(x) =cot−1(x2−x+1)*1

    Take cot−1(x2−x+1) as a first function as per ILATE, and 1 as second function. Then you can integrate by parts. :wink:
     
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