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Homework Help: Trigonometric Integral!

  1. Feb 5, 2012 #1
    1. The problem statement, all variables and given/known data

    The question asks me to evaluate the indefinite integral:


    The (cos2x)^2 is read cos squared 2x, i just dont know how to put the squared before the 2x and not have it look confusing !

    2. Relevant equations

    3. The attempt at a solution

    To be honest i have NOO clue where to start. Right now in class we are learning about inverse trig, hyperbolic trig, inverse hyperbolic trig, and the derivatives for each of them.
    ANY hint on how to start this would be greatly appreciated.
  2. jcsd
  3. Feb 5, 2012 #2


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    A substitution seems like a reasonable way to start. You pretty much have two choices: either u = sin(2x) or u = cos(2x). Only one of these will take you very far.
  4. Feb 5, 2012 #3
    How did i not see that...hhaha


    SO, i set:
    u = 1 + (cos2x)^2
    du = -4(cos2x)(sin2x) dx
    dx = du / -4(cos2x)(sin2x)

    when i insert this though, i dont seem to get far.
    is my derivative of 1+(cos2x)^2 correct?
  5. Feb 5, 2012 #4


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    That's not the substitution suggested by jbunniii.

    He suggested letting u = cos(2t) . This should give something that's related to the derivative of the arctan(u).

    NOT u = 1+cos2(2t) .
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