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Homework Help: Simple Integral

  1. Sep 1, 2010 #1
    1. The problem statement, all variables and given/known data

    Hello. I have a simple integral here that has been stumping me for the last 30 minutes. It appears that my basic integration skills have gotten very rusty.

    2. Relevant equations

    [tex]\int{x^3}\sqrt{1+x^2}dx[/tex]

    3. The attempt at a solution
    I am pretty sure a simple substitution will do, but I have yet to find it. I have tried simplifying the expression various ways, integration by parts, and I have also tried a few substitutions. Any ideas?
     
  2. jcsd
  3. Sep 1, 2010 #2

    Dick

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    Science Advisor
    Homework Helper

    u=1+x^2 will do it.
     
  4. Sep 1, 2010 #3

    jgens

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    Gold Member

    The substitution x=tan(u) also works.
     
  5. Sep 1, 2010 #4
    Thank you Dick and Jgens.

    [tex]\int{x^3}\sqrt{1+x^2}dx[/tex]

    Let u=1+x^2[tex]\rightarrow[/tex]du=2xdx

    [tex]\int{x^3}\sqrt{1+x^2}dx=\frac{1}{2}\int(u-1)\sqrt{u}du[/tex]

    [tex]\frac{1}{2}\int(u-1)\sqrt{u}du=\frac{1}{2}(\int{u^{\frac{3}{2}}}du-\int{u^{\frac{1}{2}}du)[/tex]

    [tex]\int{u^{\frac{3}{2}}}du=\frac{2}{5}u^{\frac{5}{2}}[/tex]

    [tex]\int{u^{\frac{1}{2}}du=\frac{2}{3}u^{\frac{3}{2}}[/tex]

    [tex]\frac{1}{2}\int(u-1)\sqrt{u}du=\frac{2}{10}u^{\frac{5}{2}}-\frac{2}{6}u^{\frac{3}{2}}[/tex]


    It appears I made a mistake somewhere. I believe my result should be multiplied by 1/2, and I can't find where I left that out.
     
  6. Sep 1, 2010 #5
    Nevermind. I guess the second part of the problem is incorrect. Mathematica has arrived at the same conclusion as this also. Thanks for all the help.
     
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