Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Question on an Integral
  1. Integration question (Replies: 4)

  2. Integral question (Replies: 9)

  3. Integral Question (Replies: 3)

  4. Integration question (Replies: 4)

Loading...