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Evaluating this integral

  1. Feb 28, 2012 #1
    1. The problem statement, all variables and given/known data

    I want to find the integral from 0 to 2pi of

    sin^3 t + t + cos^2 t


    2. Relevant equations



    3. The attempt at a solution

    I could convert sin^3 t to sin t (sin^2 t) and then use the identity 1 - cos 2t/2 and same for cos^2 t but I was wondering if there is a less messy way to evaluate this.
     
  2. jcsd
  3. Feb 28, 2012 #2

    Dick

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    Since your limits are 0 to 2pi if you clever you should be able to come up with an argument about the value of the sin(t)^3 part without integrating it. But you could also use another even simpler identity for sin(t)^2.
     
  4. Feb 28, 2012 #3
    Ah, I see that it is 0 for sin^3 t, does this hold for any period of 2pi for sin^3? Also what is the other identity that I could use? I thought it would just be the half angle identity.
     
  5. Feb 28, 2012 #4

    Curious3141

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    Integrating the t term is trivial.

    The cos^2 term can be handled with the double angle formula for cosine.

    The sin^3 term can be immediately reduced to zero by sketching the curve for y = sin^3x and observing that it's an odd periodic function. It applies for any *even* multiple of pi at both upper and lower bounds.
     
  6. Feb 28, 2012 #5

    Dick

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    sin(t)^2=1-cos(t)^2.
     
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