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Thread: Trig integration

  1. #1
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    Trig integration

    I'm having a bit of trouble integrating:

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

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

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

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

    $\displaystyle sin^{2}(2x)+cos^{2}(2x)+2sin(2x)cos(2x)$

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