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Math Help - Indefinite Integral

  1. #1
    Member Em Yeu Anh's Avatar
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    Angry Indefinite Integral

    Evaluate \int{x^3\sqrt{x^2+36}} dx

    Stuck on this one.
    Last edited by Em Yeu Anh; November 26th 2009 at 08:49 PM. Reason: Forgot the "dx" on the integral, hehe =)
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Em Yeu Anh View Post
    Evaluate \int{x^3\sqrt{x^2+36}}\,dx

    Stuck on this one.

    Apply the substitution u=x^2+36\implies\tfrac{1}{2}\,du=x\,dx.

    Thus, \int x^3\sqrt{x^2+36}\,dx\xrightarrow{u=x^2+36}{}\tfrac  {1}{2}\int (u-36)\sqrt{u}\,du.

    Can you take it from here?
    Last edited by Chris L T521; November 26th 2009 at 08:00 PM.
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  3. #3
    Member Em Yeu Anh's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    Apply the substitution u=x^2+36\implies\tfrac{1}{2}\,du=\,dx.

    Thus, \int x^3\sqrt{x^2+36}\,dx\xrightarrow{u=x^2+36}{}\tfrac  {1}{2}\int (u-36)\sqrt{u}\,du.

    Can you take it from here?
    I'm slightly confused on the last part that has (u-36); is that supposed to represent the x^3?
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  4. #4
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    i hope it can help you :

    u=x^2+36\implies\tfrac{1}{2}\,du=x\,dx

    x^2=u-36

    \int x^3\sqrt{x^2+36}\,dx=\int x^2\sqrt{x^2+36}\,xdx=\int (u-36)\sqrt{u}\,\tfrac{1}{2}du =\tfrac{1}{2}\int (u-36)\sqrt{u}\,du
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  5. #5
    Member Em Yeu Anh's Avatar
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    Anyone mind checking the complete solution before I submit it?

    \frac{1}{2}\int(u-36)\sqrt{u} = \frac{1}{2}\int{u^{\frac{3}{2}}-36u^{\frac{1}{2}}}

     = \frac{1}{2}(\frac{2}{5}u^{\frac{5}{2}}-24u^{\frac{3}{2}})+ C = \frac{2}{10}(x^2+36)^{\frac{5}{2}} - 12(x^2+36)^{\frac{3}{2}} + C
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  6. #6
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    Quote Originally Posted by Em Yeu Anh View Post
    Anyone mind checking the complete solution before I submit it?

    \frac{1}{2}\int(u-36)\sqrt{u} = \frac{1}{2}\int{u^{\frac{3}{2}}-36u^{\frac{1}{2}}}

     = \frac{1}{2}(\frac{2}{5}u^{\frac{5}{2}}-24u^{\frac{3}{2}})+ C = \frac{2}{10}(x^2+36)^{\frac{5}{2}} - 12(x^2+36)^{\frac{3}{2}} + C
    Check your answer at this website: Wolfram|Alpha
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