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Need help with this integral

  1. Aug 20, 2014 #1
    Hey. I'm new to the forum and I was hoping you could help me solve this integral. I was searching for a clue on similar integrals posted on internet, but I couldnt find anything helpful.

    The integral is in attachment

    Attached Files:

  2. jcsd
  3. Aug 20, 2014 #2
    I have not worked this out but the substitution u = cos(x) followed by partial fractions seems like it could do the trick.
  4. Aug 21, 2014 #3
    does t=tan(x/2) work in this case?
  5. Aug 21, 2014 #4


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    What's wrong with u = cos(x) as suggested above?
  6. Aug 21, 2014 #5
    Nothing really. I just have difficulties recognizing what I should take as a substitution.

    EDIT: If I take u=cosx, du=-sinx dx which makes dx=-du/sinx ... and everywhere else there's "u" instead of "cosx"
    Last edited: Aug 21, 2014
  7. Aug 21, 2014 #6
    You already have sin x dx under the integral. That becomes -du, all your cosines become u, and you get integral of [itex]\frac{4(u-1)}{u^2(2-u)}[/itex]
  8. Aug 21, 2014 #7
    wow...just WOW... I didnt even notice there's "sinx" in the numerator, how stupid of me. Thank you and I'm really sorry for wasting your time.
  9. Aug 31, 2014 #8
    If you do integration by substitution than partial fractions you can get the the right answer

    You also had sin(x)sin(x) at the bottom
    (which is cos^2(x) which will help you greatly)
  10. Sep 14, 2014 #9


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    Try multiplying the numerator and denominator by [sec(x)]^2, then do the substitutions.
  11. Sep 14, 2014 #10


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    You might need to substitute twice, meaning you'll want to back-substitute twice into your final answer.
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