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Question on an Integral

  1. Feb 16, 2010 #1
    Hi guys, I have a quick question.
    When integrating [tex]\int sin^2xdx[/tex], do you need to use the trig substitution [tex]\frac{1-cos(2x)}{2}[/tex]?
    I was able to integrate it using just [tex]1-cos^2x[/tex] as my identity and even when I showed a calc teacher how I solved the problem, he doesn't believe me or my math for some reason even though I proved that the two methods get the same result but in a different form.
  2. jcsd
  3. Feb 16, 2010 #2
    The best way to know is for you to post your math.
  4. Feb 16, 2010 #3
    That looks like a textbook integration by parts problem. Could that be how your teacher wants you to solve it?
  5. Feb 16, 2010 #4
    Yeah, I did it using integration by parts even though I was supposed to use the trig substitution.
    [tex]\int sin^2xdx[/tex]
    [tex]Let u = sinx, dv = sinxdx[/tex]
    [tex]du = cosxdx, v = -cosx[/tex]
    [tex]-sinxcosx + \int cos^2xdx[/tex]
    [tex]\int sin^2xdx = -sinxcosx + \int 1 - sin^2xdx[/tex]
    [tex]\int sin^2xdx = -sinxcosx + x - \int sin^2xdx[/tex]
    [tex]2\int sin^2xdx = -sinxcosx + x[/tex]
    [tex]\int sin^2xdx = \dfrac{1}{2}x - \dfrac{1}{2}sinxcosx[/tex]

    Both should work right? You don't "have" to use the cos(2x) thing right?
  6. Feb 16, 2010 #5
    They both definitely work. But whether or not you have to use Trig.Sub. is up to who ever controls your grade.
  7. Feb 17, 2010 #6
    Okay cool. And this wasn't for a grade or anything just to be clear.
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