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Another Integration Question :S

  1. Dec 25, 2011 #1
    1. The problem statement, all variables and given/known data

    ∫(sin(x-2))^(1/3)

    3. The attempt at a solution

    I reached until -∫1/((1-u^2)^1/3).du where u = sin(x)
    Then i got stuck as i tried to integrate to inverse function but it doesn't work as that is for ∫1/((a-x^2)^1/2)

    another try was to seperate the sin(x-2)

    where i reached until

    ∫(1-(cos(x-2))^2)*(sin(x-2))^(5/3).dx
    then using u= cos(x-2) u get -du=sin(x-2) . but this doesn't work too -.- !

    Help
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 25, 2011 #2
    Do you know this or no?

    NumberedEquation3.gif
     
  4. Dec 25, 2011 #3
    Yes , this is the function of integration .
    If u mean ∫(sin(x-2))^(1/3).dx = 3/4*cos(x-2)^(4/3) + C . it is wrong :S
     
  5. Dec 25, 2011 #4

    SammyS

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    Hello Redoctober.

    WolframAlpha leads me to believe that this can't be expressed in terms of elementary functions.

    You might try a power series method.

    By The Way: You should be posting these integration questions in the Calculus and Beyond section.
     
  6. Dec 25, 2011 #5
    Oh k thanks :) . I think i ll keep it for later , as we didn't start yet with the power series method yet in the Calculus 1 course .

    K i ll post in the Calculus and Beyond :) .
     
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