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Evaluating an Indefinite Integral using Substitution

  1. Sep 4, 2015 #1

    Drakkith

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    1. The problem statement, all variables and given/known data
    Evaluate the Integral:
    ∫sin2x dx/(1+cos2x)

    2. Relevant equations


    3. The attempt at a solution

    I first broke the numerator up:
    ∫2sinxcosx dx /(1+cos2x)
    2∫sinxcosx dx /(1+cos2x)

    Then I let u = cosx so that du = -sinx dx
    -2∫u du/(1+u2)

    And now I'm stuck. I thought about turning 1+u2 into tan-1u + c, but I still have another u to deal with in the numerator.
     
  2. jcsd
  3. Sep 4, 2015 #2

    SammyS

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    Try a further substitution.

    What is the derivative of 1 + u2 ?
     
  4. Sep 4, 2015 #3

    Drakkith

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    Okay. So if I let y = 1+u2, then dy = 2u du, or dy/2 = u du

    That gives me -2/2∫y-1 dy
    -∫y-1 dy

    Integrating:
    -ln|y| + c

    Since y = 1+u2 that becomes:
    -ln|1+u2|+ c

    And since u = cosx:
    -ln |1 + cos2x| + c
     
  5. Sep 4, 2015 #4
    hint: differentiation of ## ln(x) = \frac{1}{x} ##
     
  6. Sep 4, 2015 #5

    SammyS

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    Right.

    So looking back at your two substitutions, letting u = 1 + cos2(x) would have worked very nicely.
     
  7. Sep 4, 2015 #6

    epenguin

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    Er, before you do any calculus can't you very significantly simplify the thing you are asked to integrate?
     
  8. Sep 4, 2015 #7

    Drakkith

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    How so? I'm not getting anything that looks useful.

    u = 1 + cos2x, du = -2sinx dx
    That still leaves cosx in the numerator.

    I'm open to suggestions.
     
  9. Sep 4, 2015 #8

    SammyS

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    Check your differentiation.
     
  10. Sep 4, 2015 #9

    Drakkith

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    Oh. Apparently it's -2sinxcosx. Yeah that would work nicely.

    It's been a while since I did any differentiation and I appear to have forgotten some important details about how to use the chain rule.
     
  11. Sep 4, 2015 #10

    epenguin

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    Not only?
     
  12. Sep 4, 2015 #11

    Drakkith

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    Huh?
     
  13. Sep 4, 2015 #12

    epenguin

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    OK my late night mistake..
     
  14. Sep 4, 2015 #13

    Drakkith

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    No worries!

    Anyways, thanks for the help all! Sammy, I think I'll have to spend a little time going of differentiation rules again...
     
  15. Sep 4, 2015 #14

    SammyS

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    Sure. Practice helps.

    It was interesting that you got the part of the chain rule that most people miss and missed the part they get.
     
  16. Sep 5, 2015 #15

    epenguin

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    Conclusion?
     
  17. Sep 5, 2015 #16

    SammyS

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    This was the problem Drakkith was asking about:
    This was his final result.
    After that I suggested a single substitution to do the job, which he indicates he now adequately understands.
     
  18. Sep 5, 2015 #17

    epenguin

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    Ah yes, I had not read carefully enough, I just read 'substitution' and thought oof, that is a bit heavy. In fact in #1 he was only missing that the numerator is (apart from numerical factor) is the derivative of the denominator. I had just mentally written
    [tex] 2 \int \frac{cos x sin x}{1 + cos^2 x} dx [/tex]
    [tex]. = - 2 \int \frac{cos x }{1 + cos^2 x} d (cos x) [/tex]
    [tex]. =-. \int \frac{d (cos^2 x)}{1 + cos^2 x}. [/tex]
    [tex]. = - \int \frac{d (1 + cos^2 x)}{1 + cos^2 x}. [/tex]
    [tex]. = - ln (1 + cos^2 x) + C [/tex]
    which you can also write
    [tex]. = - ln K (1 + cos^2 x)[/tex]

    which is just the some thing as Drakkith, just slightly differenti setting out.
     
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