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Definite Integral of Trig Function

  1. Feb 13, 2012 #1
    1. The problem statement, all variables and given/known data
    ∫sin2xcosx x=0,pi/4


    2. Relevant equations



    3. The attempt at a solution
    By double angle magic:
    ∫sin2xcosx=
    ∫2sinxcos^2x
    u=cosx
    du=-sinxdx
    -2∫u^2
    -2(u^3)/3
    -2cos^3x/3
    = -1/(3√2) for x=pi/4

    How does that look folks?
     
  2. jcsd
  3. Feb 14, 2012 #2

    tiny-tim

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    Science Advisor
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    Hi Jimbo57! :smile:

    (have a pi: π and try using the X2 button just above the Reply box :wink:)

    Fine down to …
    … what about for x = 0 ? :wink:
     
  4. Feb 14, 2012 #3
    Lol woops, I'm used to ignoring the 0 with non trig integrals.

    New answer:

    (-1+2√2)/3√2

    Thanks for the help Tiny-Tim! (and the pi)
     
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