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What am I doing wrong with this simple integration?

  1. Oct 2, 2013 #1
    1. The problem statement, all variables and given/known data

    $$\int { { sin }^{ 2 }x{ cos }^{ 2 }xdx }$$

    2. Relevant equations



    3. The attempt at a solution

    $$\int { { sin }^{ 2 }x{ cos }^{ 2 }xdx } \\ =\frac { 1 }{ 4 } \int { (1-cos2x)(1+cos2x)dx } \\ =\frac { 1 }{ 4 } \int { -{ cos }^{ 2 }x+1dx } \\ =\frac { x }{ 4 } -\frac { 1 }{ 8 } \int { cos2x+1 } dx\\ =\frac { x }{ 4 } -\frac { 1 }{ 8 } [\frac { sin2x }{ 2 } +x]\\ =\frac { 2x-sin2x }{ 16 }$$

    It's incredibly frustrating to do so many stupid mistakes on such simple integration. I can't believe how weak I've gotten at doing integrals after only 4 months...

    Edit: never mind. I got it. I put x instead of 2x on line 3. Really dumb mistakes but I keep making them from a result of being so rusty.
     
  2. jcsd
  3. Oct 2, 2013 #2

    tiny-tim

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    Hi Turion! :smile:

    Even quicker would have been sin2xcos2x = (1/4)sin22x = … ? :wink:
     
  4. Oct 2, 2013 #3

    jfizzix

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    I think the problem is in the difference of two squares
    [itex](1-Cos(2x))(1+Cos(2x)) = 1-Cos^{2}(2x)[/itex]
     
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