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Integrating Trig

  1. Sep 7, 2005 #1
    Ok, so we have

    [itex]\int_{0}^{1}\left(\sin{2x}*\cos{2x}\right)dx[/itex]

    Using the double angle forumla we change the integrand

    [itex](1/2)\int_{0}^{1}\left(2*\sin{2x}*\cos{2x}\right)dx[/itex]

    which converts to

    [itex](1/2)\int_{0}^{1}\left(\sin{4x}\right)dx[/itex]

    This is where I run into trouble... I'm trying to use the formula
    [itex]\int\left(\sin{bx}\right)dx = (1/-b)\cos{bx}[/itex]

    but my answers are not working. I'm thinking that it has something to do with the fact this is a deffinite integral... any help?

    Thx
     
  2. jcsd
  3. Sep 7, 2005 #2
    How are you answers not working?

    [tex]-\frac{1}{4}\cos{4x}\bigvert|_0^1[/tex]

    As long as you know the value of cos(0), you shouldn't have any trouble!
     
  4. Sep 7, 2005 #3
    .... *sigh* I was so used to seeing a 0 and just subtracting by 0 that I forgot cos(0) is in fact NOT 0... Is it bad that I take shortcuts without even realizing I'm taking them? :cry:

    EDIT: I meant WRONG shortcuts =)
     
  5. Sep 8, 2005 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    By the way, I would not have worried about combining those trig functions.

    To integrate [tex]\int_0^1 sin(2x)cos(2x)dx[/tex] just note that the sine and cosine are both to an odd exponent (1). Let u= sin(2x) so that du= 2 cos(2x)dx or (1/2)du= cos(2x)dx. When x= 0, u= sin(0)= 0, when x= 1, u= sin(2). The integral becomes
    [tex]\frac{1}{2}\int_0^{sin(2)}udu[/tex]
    which equals
    [tex]\frac{1}{4}u^2[/tex] evaluated between 0 and sin(2) and is
    [tex]\frac{1}{4}sin^2(2)[/tex]
     
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