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How to integrate this

  1. Feb 3, 2013 #1
    [tex]\displaystyle\int sin^22tcos^2t\ dt [/tex]

    This was part (b) to a question, the previous part of the question was to integrate [itex] \displaystyle\int sin^22tcost\ dt [/itex] which I managed to do by expressing [itex] sin^22t [/itex] as [itex] 4(sin^2t - sin^4t) [/itex]

    I tried a similar method for the integrand above, but didn't really go far with it. I'm not really sure what I'm trying to spot here, and I usually struggle with these integrals.
     
  2. jcsd
  3. Feb 3, 2013 #2
    You can try with [itex]\int sin^2(2t)dt - \int sin^2(2t)sin^2(t)dt[/itex], plus some trigonometric manipulations.

    P.S.: Silly me, perhaps you only need some trigonometric manipulations.
     
  4. Feb 3, 2013 #3
    thanks
     
  5. Feb 3, 2013 #4

    vela

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    Instead of changing the first factor, I'd use the identity ##\cos^2 t = \frac{1+\cos 2t}{2}## to get everything in terms of ##2t##. It should be pretty straightforward after that.

    Your original approach would also work. You should check your textbook on how to handle integrals of the form
    $$\int \cos^n x \sin^m x\,dx.$$ They're tedious, but there's a recipe to follow that depends on the evenness and oddness of ##m## and ##n##.
     
  6. Feb 3, 2013 #5

    haruspex

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    After following vela's advice, you may also find it useful to know cos(3x) = 4 cos3(x) - 3 cos(x)
     
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