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Integral from -pi/3 to 0 of 3sin^3(x)dx ?

  1. Feb 19, 2013 #1
    ∫from [itex]-\pi[/itex]/3 to 0 of 3sec3xdx

    Integration by parts...

    3∫sec2x*secxdx

    dv= 3sec2x
    v= 3tanx

    u= secx
    du= secxtanx

    Formula: uv - ∫vdu

    secx*3tanx - ∫3tanx*secxtanxdx

    Integration by parts again...

    u= 3tanx
    du= 3sec2x

    dv= sextanx
    v= secx

    3tanx*secx - ∫secx*3sec2x

    Altogether:

    ∫3sec3x = 3secxtanx - 3secxtanx - ∫3sec3x

    *Add ∫3sec3x on both sides*

    ∫2*3sec3x = 0 !!!

    It seems that the two 3secxtanx's would cancel! :/

    ∫6sec3x = 0

    This is like the same thing we started out with! :confused: ugh, man. I don't really know what to do now.
    I know this is a definite integral, so we should substitute -pi/3 and 0 in the end, but we still have an integral! #=_=

    ...This definitely does not seem right! :(
    But I don't really see what's wrong with my work.
    Please help! Thank you so much! :D
     
    Last edited by a moderator: Feb 19, 2013
  2. jcsd
  3. Feb 19, 2013 #2

    Mark44

    Staff: Mentor

    That's not the way to go. Your integral above can be written as
    ## \int tan^2(x) sec(x) dx##
    ## = \int (sec^2(x) - 1) sec(x) dx ##
    This will give you another sec3 integral and one that's somewhat easier. You should be able to solve algebraically for the integral you're looking for.
    I am not including the factor of 3 from the first integral. It would have been better to bring it out of the integral right away and just concentrate on ∫sec3(x) dx.
     
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