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Simple derivative problem

  1. Feb 4, 2004 #1
    I am taking basic calculus, and have just got to integration. Can someone please tell me how to find the antiderivative of (sin(x))^4?
     
  2. jcsd
  3. Feb 4, 2004 #2

    matt grime

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    Looks like home work. Replace a power with a multiple - you can do sin^2 using cos 2, so this is no harder.
     
  4. Feb 4, 2004 #3
    Sorry, but I still don't understand. How do you find the antiderivative of (sin(x))^2? Could you explain it to me please?
     
  5. Feb 4, 2004 #4

    NateTG

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    Do you know about the chain rule?
    do you know the derivatives of [tex]\sin x[/tex] and [tex]x^2[/tex]?
     
  6. Feb 5, 2004 #5
    Yes, but how are they going to help me find the antiderivative of [tex]sin^4x[/tex]?
     
  7. Feb 5, 2004 #6
    Try to Reduce the power of sin4x by
    2sin2x=1-cos2x.

    therefore
    4sin4x=(1-cos2x)2

    i.e 1+cos22x-2cos2x
    Again use

    2cos22x = 1+cos24x

    Simplifying u will obtain

    [tex]\sin^4x = \frac{3}{8} +\frac{cos4x}{8}-\frac{cos2x}{2}[/tex]

    Hope this will help u
     
  8. Feb 5, 2004 #7

    Integral

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    Himanshu

    I think you have a typo

    Should be

     
  9. Feb 5, 2004 #8
    Ya typo is there it is

    2cos22x = 1+cos4x

    error is regretted
     
  10. Feb 5, 2004 #9
    Himanshu

    Thanks a lot, I understand now.
     
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