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Silly u-substitution mistake happening somewhere

  1. Sep 17, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\int[/tex][tex]\frac{sin(2x)}{1+cos(x)^2}[/tex]


    2. Relevant equations
    None?



    3. The attempt at a solution

    I know I can use a trig identity to end up with a numerator of -- 2sin(x)cos(x)

    So:


    [tex]\int[/tex][tex]\frac{2sin(x)cos(x)}{1+cos(x)^2}[/tex]



    I am using u=1+cos(x)^2 and du=-2sin(x)dx

    Substitute in and I end up with

    [tex]\int[/tex][tex]\frac{-cos(x)}{u}[/tex]du

    And that's where I hit a wall, because I still have a cos(x) in there. Anyone willing to offer hints on this one? Thanks much!
     
  2. jcsd
  3. Sep 17, 2009 #2

    Kurdt

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    Staff Emeritus
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    Gold Member

    Try [itex]u=cos^2(x)[/itex].
     
  4. Sep 17, 2009 #3
    Oh man, that would make sense... Gah, thankyou :)
     
  5. Sep 17, 2009 #4
    If [itex]u=1+\cos^2(x)[/itex] then [itex]du \neq -2\sin(x)\;dx[/itex]

    It would be [itex]du=-2\sin(x)\cos(x)\;dx[/itex]

    --Elucidus
     
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