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Integral of trigonometric function

  1. Jun 19, 2008 #1
    How do I find the derivative of cos2x?
    Do I use the chain rule?
    u=2x
    u'=2
    y=cosu
    y'=-sinu
    dy/dx=-2sin(2x)
    Edit: Sorry. I mean derivative, not integral.
     
    Last edited: Jun 19, 2008
  2. jcsd
  3. Jun 19, 2008 #2
    That looks good, my friend, so I think you've got it!
     
  4. Jun 19, 2008 #3
    yes, that's correct

    Now try it all in one step!
     
  5. Jun 19, 2008 #4
    what about the integral of xsin(x/2).
    How would I do that?
     
  6. Jun 19, 2008 #5
    Integral or Derivative? You might want to write it out properly ...
     
  7. Jun 19, 2008 #6
    integral of xsin(x/2) dx
     
  8. Jun 19, 2008 #7
    [tex]\int x\sin{\frac x 2}dx[/tex]

    Try integrating by parts.
     
  9. Jun 19, 2008 #8
    use sin(x/2) identity and try to convert everything to sin..

    I have a feeling that there's an easier way to do that
     
  10. Jun 20, 2008 #9

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi Ry122! :smile:

    You can also use the chain rule without having to define a u:

    dcos2x/dx = (dcos2x/d(2x))(d(2x)/dx) = (-sin2x)(2). :smile:

    (That way, you can eventually do these things in your head! :wink:)
     
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