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Math Help - Trig integration

  1. #1
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    Trig integration

    I'm having a bit of trouble integrating:

    \int(sin(2x) + cos(2x))^2 \ dx. According to the solution in the back of my book, it's supposed to be  x - (1/4)cos(4x) + C.

    I was thinking of using integration by parts with my  dv = (sin(2x) + cos(2x)). Anybody have any ideas on how to solve this?
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  2. #2
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    Use the identity (sin(2x)+cos(2x))^{2}=sin(4x)+1

    Then \int\left[sin(4x)+1\right]dx

    Easier now?.
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  3. #3
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    thanks, but how did you simplify it to \int\left[sin(4x)+1\right]dx ?
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  4. #4
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    Expand it out and you can see it.

    sin^{2}(2x)+cos^{2}(2x)+2sin(2x)cos(2x)

    If you know your identites you can see that sin^{2}(2x)+cos^{2}(2x)=1 and since sin(2x)=2sinxcosx then
    2sin(2x)cos(2x)=sin(4x)
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