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Trigonometric Integral 2

  1. May 23, 2009 #1

    Mentallic

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    Part 2, leading on from https://www.physicsforums.com/showthread.php?t=315803"

    1. The problem statement, all variables and given/known data
    Given [itex]f(x)=cotx+tanx[/itex] find
    [tex]\int{[f(x)]^2}dx[/tex]

    3. The attempt at a solution
    I've attempted many different varieties of approaches to the problem. Trying to use the substitution method for sinx, cosx, tanx.... and a few others... re-arranging the function, trying to get it into a more convenient form... trying to use some of the ideas given in the first thread... No luck.

    Basically, it has all been a bunch of guessing and hoping something useful will appear. Plus another bunch of frustration on my part, but I wont get into the details of that xD
     
    Last edited by a moderator: Apr 24, 2017
  2. jcsd
  3. May 24, 2009 #2

    rock.freak667

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    Expand out [f(x)]^2

    use [itex]cot^2x+1=cosec^2x[/itex] and [itex]tan^2x+1=sec^2x[/itex]
     
  4. May 24, 2009 #3

    Mentallic

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    Thanks rockfreak :smile:

    It turns out to be tanx-cotx. I also obtained [itex]\int{sec^2x+cosec^2x}dx[/itex] through another longer method but it strike me at that moment that I can take the integral of each of these. (I think I'll keep my standard integral formulas close-by next time).

    Thanks again.
     
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