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Thread: Indefinite Integral

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

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

    Stuck on this one.
    Last edited by Em Yeu Anh; Nov 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 $\displaystyle \int{x^3\sqrt{x^2+36}}\,dx$

    Stuck on this one.

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

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

    Can you take it from here?
    Last edited by Chris L T521; Nov 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 $\displaystyle u=x^2+36\implies\tfrac{1}{2}\,du=\,dx$.

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

    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 :

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

    $\displaystyle x^2=u-36$

    $\displaystyle \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$$\displaystyle =\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?

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

    $\displaystyle = \frac{1}{2}(\frac{2}{5}u^{\frac{5}{2}}-24u^{\frac{3}{2}})+ C = \frac{2}{10}(x^2+36)^{\frac{5}{2}} - 12(x^2+36)^{\frac{3}{2}} + C$
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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?

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

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