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Math Help - integral (w^5)(squareroot((w^3)-1)

  1. #1
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    integral (w^5)(squareroot((w^3)-1)

    u= w^3
    du= 3w^2dw

    integral 1/3[u (u-1)^1/2 du

    1/3 [ u^3/2 -udu]
    1/3 [[ 2/5(u^5/2) - 1/2u^2]]

    the answer is

    1/3[2/5(u^5/2) - 2/3(u^3/2)

    what did I do wrong in the second part?
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  2. #2
    o_O
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    \int w^{5} \sqrt{w^{3} - 1}dw = \int w^{3} w^{2} \sqrt{w^{3} - 1} dw

    u = {\color{red}w^{3} - 1} \: \: \Rightarrow \: \: du = 3w^{2}dw \: \: \Rightarrow \: \: {\color{blue}w^{2}dw = \frac{1}{3}du}
    u + 1 = {\color{magenta}w^{3}}

    Make your subs:
    \int {\color{magenta}w^{3}} {\color{blue}w^{2}} \sqrt{{\color{red}w^{3} - 1}} {\color{blue}dw}
    = \int {\color{magenta}(u+1)}{\color{red}u^{\frac{1}{2}}} {\color{blue}\frac{1}{3}du}

    Seems like you messed up dealing with w^5.
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  3. #3
    Math Engineering Student
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    Substitution z^2=w^3-1 will kill quickly the problem.
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