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Indefinite Integration by u-sub/trig sub

  1. Apr 13, 2007 #1
    1. The problem statement, all variables and given/known data
    Integrate (x^3)sqrt(1-x^2)


    2. Relevant equations



    3. The attempt at a solution

    I used trig. substitution along with u substitution and came up with (x^4)/4 +C which I know is wrong. My professor gave the answer -(((3x^2)+2)((1-x^2)^(3/2)))/15 . Please help!!!!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 13, 2007 #2
    1. The problem statement, all variables and given/known data
    Integrate (x^3)sqrt(1-x^2)


    2. Relevant equations



    3. The attempt at a solution

    I used trig. substitution along with u substitution and came up with (x^4)/4 +C which I know is wrong. My professor gave the answer -(((3x^2)+2)((1-x^2)^(3/2)))/15 . Please help!!!!
     
  4. Apr 13, 2007 #3

    quasar987

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    Is there a special restriction in the question that says you can't use integration by parts?
     
  5. Apr 13, 2007 #4

    Dick

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    The substitution you want is u=1-x^2. One x from the x^3 will then be used to create the du leaving x^2 but then you'll remember that x^2=1-u. Try it, you'll like it!
     
  6. Apr 14, 2007 #5

    Gib Z

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    Which substitution? x=sin theta should work fine.
     
  7. Apr 14, 2007 #6

    Gib Z

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    Ok I worked through it for you.
    [tex]\int \frac{x^3}{\sqrt{1-x^2}} dx[/tex]
    x=sin theta
    dx=cos theta d(theta)
    [tex]\int \frac{x^3}{\sqrt{1-x^2}} dx = \int \sin^3 \theta d\theta[/tex]
    [tex]=\int \sin^2 \theta \cdot \sin \theta d\theta = \int (1-\cos^2 \theta)\sin \theta d\theta = \int \sin \theta d\theta - \int \cos^2 \theta \sin \theta d\theta[/tex]

    First integral is just -cos theta, 2nd integral is simple with u=cos theta. Easy as pi.
     
  8. Apr 14, 2007 #7
    Yep - I get the Prof's result too.

    First substitution: [tex]u= x^2[/tex]

    Second substition: let [tex]v^2 = 1-u[/tex]

    You should wind up with an integral of:

    I = [tex]\int (v^4-v^2) dv[/tex]
     
    Last edited: Apr 14, 2007
  9. Apr 14, 2007 #8
    Only if it were that simple. The radical is in the numerator. :wink:
     
  10. Apr 14, 2007 #9

    Gib Z

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    I can't read. Kill me. I'll get another solution for you tomorrow then, for the right problem :P
     
  11. Apr 14, 2007 #10

    Gib Z

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    Ok actually no problem either way.
    Same substitution, x=sin theta
    dx= cos theta d(theta)

    [tex]\int x^3 \sqrt{1-x^2} dx = \int \sin^3 \theta \cos^2 \theta d\theta = \int \sin \theta (1-\cos^2 \theta)\cos^2 \theta d\theta = \int (\cos^2 \theta - \cos^4 \theta) \sin \theta d\theta[/tex] Which is very simple with the easy substitution, u=cos x.

    EDIT: Rewritten in the form [tex]\int (\cos^4 \theta - \cos^2 \theta) (-\sin \theta) d\theta[/tex] it becomes what TheoMcCloskey said it would be.
     
    Last edited: Apr 14, 2007
  12. Apr 14, 2007 #11

    HallsofIvy

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    But as Dick and ThomasMcCloskey (indirectly) said, The substitution u= 1-x2 is simpler.
     
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