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Calculus 2 Integral

  1. May 18, 2005 #1
    How would you find the integral of x^3 * ( x^2 - 8 )^1/2 ?

    I've tried everything I can think of; substitution / trigonometric substitution, partial fractions, integration by parts, but I just can't get the answer. The answer is supposed to be:

    1/15(x^2-8)^3/2 * (3x^2 + 16 )
     
  2. jcsd
  3. May 18, 2005 #2
    Separate x^3 into x^2*x, then do integration by parts with x^2 for one and x(x^2-8)^.5 for the other.
     
  4. May 18, 2005 #3

    dextercioby

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    You could try a hyperbolic substitution.

    [tex] \frac{x}{2\sqrt{2}}=\cosh t [/tex]

    Daniel.
     
  5. May 19, 2005 #4
    Or just let u = x2 - 8, so x^3dx = (u+8)du/2.

    --J
     
  6. May 19, 2005 #5
    Or you can do trigonometric substitution.

    [tex] x = \sqrt{8} sec\theta; [/tex]

    [tex] dx = \sqrt{8} sec\theta tan\theta d\theta ;[/tex]

    and [tex] x^3 = 8\sqrt{8} sec^3\theta [/tex]

    Edit message:

    Blah i would definitely go for Justin's substitutions.
     
    Last edited: May 19, 2005
  7. May 19, 2005 #6

    GCT

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    Trig substitution will definitely work, you can simplify towards the step

    [tex]64 \sqrt{(8)} \int sec^{4} \theta~tan^{2} \thetad \theta [/tex]

    [tex]64 \sqrt{(8)} \int sec^{2} \theta~(1+tan^{2} \theta )tan^{2} \theta d \theta [/tex]

    [tex]u=tan \theta [/tex]

    [tex]du= sec^{2} \theta d \theta [/tex]
     
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