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Two (equivalent?) integrals

  1. Oct 7, 2008 #1
    1. This is part of a larger calculation. I do it one way and get the right answer and do it a seemingly equivalent way and get the wrong answer. The question is just about reconciling the two integrals below.

    2.

    [tex]I_1 = \int \frac{sin(2x)}{2} dx = -\frac{cos(2x)}{4} [/tex]

    [tex] \frac{sin(2x)}{2} = sin(x)cos(x) [/tex]

    [tex] I_2 = \int sin(x)cos(x) dx = - \frac{cos^2(x)}{2} [/tex]

    For [tex] x= 0 [/tex], [tex] I_1 = -1/4 \neq I_2 = -1/2 [/tex].

    How can this be?
     
  2. jcsd
  3. Oct 7, 2008 #2

    Dick

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    cos(2x)=cos(x+x)=cos(x)^2-sin(x)^2=2*cos(x)^2-1. Your two expressions differ by a constant. Don't forget the +C on an indefinite integral.
     
  4. Oct 7, 2008 #3
    Thanks!
     
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