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Math Help - integration question no.5

  1. #1
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    integration question no.5

    integration question no.5
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  2. #2
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    You could then let {\theta}=sin^{-1}(u), \;\ d{\theta}=\frac{1}{\sqrt{1-u^{2}}}du

    Then making the subs, it boils down to:

    2\int\frac{u}{(u^{2}+3)^{\frac{1}{3}}}

    May as well finish it:

    If you let w=u^{2}+3, \;\ dw=2udu

    That turns into an easy integral: \int\frac{1}{w^{\frac{1}{3}}}dw=\frac{3}{2}w^{\fra  c{2}{3}}

    Resub: \frac{3}{2}(u^{2}+3)^{\frac{2}{3}}

    Resub: \frac{3}{2}(sin^{2}{\theta}+3)^{\frac{2}{3}}
    Last edited by galactus; October 30th 2007 at 07:00 AM.
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  3. #3
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    Use u = sinT?
    [Let me write T as theta---for less typing.]

    Okay.

    u = sinT
    So,
    du = cosT dT
    dT = du/cosT

    INT.[sin(2T) / cbrt(sin^2(T) +3)]dT
    = INT.[2sinTcosT *(sin^2(T) +3)^(-1/3)]dT
    Substitutions,
    = INT.[2u*cosT*(u^2 +3)^(-1/3)](du/cosT)
    = INT.[2u*(u^2 +3)^(-1/3)]du
    = INT.[(u^2 +3)^(-1/3)](2u du)

    d(u^2 +3) = 2u du

    = INT.[(u^2 +3)^(-1/3)]d(u^2 +3)
    = 1/(-1/3 +1) * (u^2 +3)^(-1/3 +1) +C
    = (3/2)(u^2 +3)^(2/3) +C

    Going back to u = sinT

    = (3/2)(sin^2(T) +3)^(2/3) +C -----------------answer.
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