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Math Help - Integration by parts

  1. #1
    Junior Member Greymalkin's Avatar
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    Integration by parts

    Integration by parts-trig2.pngHow is it that the top becomes the bottom?, my textbook says the 1/2nd are simple(im lost), the third uses the identities sin^2t=\frac{1-cos2t}{2} & cos^2t=\frac{1+cos2t}{2}, the last one uses the Pythagorean identity sin^2t+cos^2t=1, the last two seem doable what are the first two? I don't see how the 1/8 becomes 1/16 to start. Edit:ignore thumbnail, added new picture
    Attached Thumbnails Attached Thumbnails Integration by parts-trig2.png  
    Last edited by Greymalkin; November 8th 2012 at 11:12 AM.
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  2. #2
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    Re: Integration by parts

    Hi Greymalkin,

    when f(x)= cos(4x) => F(x)= 1/4* sin(4x) +C
    g(x)= sin^2(2t)*cos2t => g(x)= sin(2t))^2*(sin(2t))' *1/2 => G(x)= 1/6*sin^3(2t) +C
    Prove the results with derivation!
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  3. #3
    Junior Member Greymalkin's Avatar
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    Re: Integration by parts

    How does -\frac{1}{8}\int cos^3 2tdt equal
    \frac{1}{8}sin^2 2tcos2tdt -\frac{1}{8}\int cos2t dt I get the last part, it is a factor, but how does the sin2 fit into there?
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  4. #4
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    Re: Integration by parts

    Int=Integral
    cos^2(2t)=1-sin^2(2t);
    cos^3(2t)= cos^2(2t)*cos(2t)= (1-sin^2(2t))*cos(2t)= cos(2t) - sin^2(2t)*cos(2t)
    True is:
    Int cos^3(2t) dt = - Int sin^2(2t)*cos(2t)dt + Int cos(2t)dt =>
    -1/8*Int cos^3(2t) dt =1/8* Int sin^2(2t)*cos(2t)dt -1/8* Int cos(2t)dt
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