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Integration with trig

  1. Feb 3, 2014 #1
    1. The problem statement, all variables and given/known data

    ∫8sin4x dx

    2. Relevant equations



    3. The attempt at a solution
    8∫sin2xsin2x

    I'm not really sure what to do next. Maybe substitute a 1-cos2x in for one of the sin2x? Maybe both?
     
  2. jcsd
  3. Feb 3, 2014 #2

    Dick

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    That sort of integration involves using double angle identities in trig. Look them up and try and get started.
     
  4. Feb 3, 2014 #3

    ehild

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    Do you know the relations

    [tex]\sin^2(x)=\frac{1-\cos(2x)}{2}[/tex] and [tex]\cos^2(x)=\frac{1+\cos(2x)}{2}[/tex]

    ehild
     
  5. Feb 3, 2014 #4
    Ok, so I plugged in (1-cos(2x))/2 for both of my sin2x and then foiled.
    This is what I have now:
    2∫1+cos22x+2cos2x dx
     
  6. Feb 3, 2014 #5

    Dick

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    No, that's not what you get. There's a sign problem. But the important point is that you can use the double angle identities on ##\cos^2(2x)## as well.
     
    Last edited: Feb 3, 2014
  7. Feb 3, 2014 #6
    Oh, so it should be: 2∫1+cos22x-2cos2x dx
    Then: 2∫cos22x-cos2x dx
    Now could you split up your integrals and solve them individually?
     
  8. Feb 4, 2014 #7

    Dick

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    Of course you can split them up and solve them individually. That's the whole point. But now what happened to the "1" part? And a factor of 2 also disappeared. Just take it step by step.
     
    Last edited: Feb 4, 2014
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