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Integral with trig substitution

  1. Mar 19, 2008 #1
    [SOLVED] Integral with trig substitution

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

    Find [tex]\int(x^3)/\sqrt{x^2-9}[/tex]

    2. Relevant equations

    Trig substitution. sin^2 + cos^2 =1, and other things that you can figure out from that.

    Half angle formula, cos^2[tex]\theta[/tex]=(1+cos(2[tex]\theta[/tex]) )*.5

    3. The attempt at a solution

    Let x=3*sec[tex]\theta[/tex]
    so dx=3*sec[tex]\theta[/tex]*tan[tex]\theta[/tex] d[tex]\theta[/tex]

    When I substitute that in and simplify it, I got:

    27*[tex]\int(sec^4(\theta) d\theta)[/tex]

    And I don't know how to integrate that. Half angle formulas aren't seeming to work.

    Thanks!
     
  2. jcsd
  3. Mar 19, 2008 #2
    Or, you can transform some of the secants into tangents:

    [tex]27\int \sec^4 \theta d\theta = 27\int (1+\tan^2 \theta)\sec^2 \theta d\theta[/tex]

    and you can go from there.
     
  4. Mar 21, 2008 #3
    [tex]27\int (1+\tan^2 \theta)\sec^2 \theta d\theta[/tex]

    [tex]27\int \sec^2 \theta d\theta + 27\int \tan^2 \theta\sec^2 \theta d\theta[/tex]

    [tex]27*tan\theta + 9\tan^3 \theta[/tex]

    Haha sorry I kind of forgot that [tex]\int \sec^2\theta d\theta[/tex] was tan!
     
  5. Mar 21, 2008 #4
    You are almost done. Because the problem was originally in terms of x, you will need to transform [itex]\theta[/itex] back into x.
     
  6. Mar 22, 2008 #5

    dextercioby

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    There's no need for trigonometry. Part integration and an immediate substitution will do the trick.

    [tex] \int \frac{x^{3}}{\sqrt{x^{2}-9}}dx=\int x^{2}\frac{x}{\sqrt{x^{2}-9}}dx=\allowbreak x^{2}\sqrt{x^{2}-9}-2\int x\sqrt{x^{2}-9}\,dx=\allowbreak x^{2}\sqrt{x^{2}-9}-\frac{2}{3}\left( \sqrt{x^{2}-9}\right) ^{3} [/tex]

    up to a constant of integration.
     
  7. Mar 22, 2008 #6

    tiny-tim

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    … just simplify …

    Integration by parts not necessary:

    [tex] \int \frac{x^{3}}{\sqrt{x^{2}-9}}dx
    =\int x\sqrt{x^{2}-9}dx\,+\,\int \frac{9x}{\sqrt{x^{2}-9}}dx
    \,=\,\frac{1}{3}(x^{2}-9)^{3/2}\,+\,9\sqrt{x^{2}-9}
    \,=\,(\frac{1}{3}x^2\,+\,6)\sqrt{x^{2}-9}\,.[/tex]
     
  8. Mar 22, 2008 #7
    Polynomial division?
     
  9. Mar 25, 2008 #8
    Right...I always forget that!

    [tex]x=3*sec \theta \Rightarrow \theta = Sec^-1 (x/3)[/tex]

    [tex]\Rightarrow tan\theta = \sqrt{x^2-9}/3[/tex]

    [tex]9 \sqrt{x^2-9} + (x^2-9)^\frac{3}{2} /3[/tex]

    Shiny! Thanks
     
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