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    I don't know what to do!

    Anyone done one like this before?
    cheers
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    Ok

    sin^2x Cos^3x = sin^2x.CosX.(Cos^2X)
    = sin^2x.CosX.(1-sin^2X)
    = sin^2x. Cosx - sin^4x.cosx

    Then you can integrate that as the product pairs have one as the differential of the other
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    (Original post by Mysticmin)
    Ok

    sin^2x Cos^3x = sin^2x.CosX.(Cos^2X)
    = sin^2x.CosX.(1-sin^2X)
    = sin^2x. Cosx - sin^4x.cosx

    Then you can integrate that as the product pairs have one as the differential of the other

    ahh thankie!
    how did you know to do that?
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    (Original post by kimoni)
    ahh thankie!
    how did you know to do that?
    you have to split the cos^3x up into cos^2X * cosx. Then you use the identity cos^2x + sin^2x = 1, rearrange and you get cos^2x = 1 - sin^2x. Then subsitutute that in for cos^2x in the integration equation.

    When's your exam?
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    (Original post by Mysticmin)
    When's your exam?
    tomorrow! :eek:
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    (Original post by kimoni)
    tomorrow! :eek:
    lol, good luck, (p1?)
 
 
 
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Updated: June 8, 2004
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