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Integration by parts involving square root!

  1. Mar 4, 2013 #1
    1. The problem statement, all variables and given/known data
    |x3sqrt(4-x2)dx


    2. Relevant equations
    uv - | vdu


    3. The attempt at a solution
    u = x2 v = -1/3(4-x2)3/2
    du = -2xdx dv = x(4-x2)1/2

    uv - | vdu

    x2(1/3)(4-x2)3/2 - | 1/3(4-x2)3/2(2xdx)
    x2(1/3)(4-x2)3/2 +(1/3)|(4-x2)3/2(2xdx)

    u = 4 - x2
    du = -2xdx

    x2(1/3)(4-x2)3/2 - (1/3)| u3/2du
    x2(1/3)(4-x2)3/2-(1/3)2/5u5/2
    x2(1/3)(4-x2)3/2-(1/3)(2/5)(4-x2)5/2
    x2-(1/3)(4-x2)3/2-(2/15)(4-x2)5/2

    Is this all I need? The answer I was supposed to get is -1/15(4-x2)3/2(8+3x2) I guess I dont see how that correlates
     
    Last edited: Mar 4, 2013
  2. jcsd
  3. Mar 4, 2013 #2

    SammyS

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    Hello learnonthefly. Welcome to PF !

    I made a correction & did some editing above.

    That is kind of hard to read.

    What is the question that you have for us ?
     
  4. Mar 4, 2013 #3

    Curious3141

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    You didn't actually ask a question. And it's tough to read what you've written - please format in Latex, it makes things a lot easier. :smile:

    At any rate, your final answer is wrong, but it's probably a simple sign error.

    For example, in the first integration by parts, ##v = -\frac{1}{3}{(4-x^2)}^{\frac{3}{2}}##. This is because of the ##-x^2## term within the parentheses.

    You should go through your working carefully looking at all the sign errors and correct them.

    EDIT: beaten by SammyS :biggrin:
     
  5. Mar 4, 2013 #4
    Sorry guys, updated and cleaned
     
  6. Mar 4, 2013 #5

    SammyS

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    Your result is equivalent to:

    [itex]\displaystyle (1/15) (4-x^2)^{3/2} (3 x^2+8)[/itex]

    You have a sign error in you integration by substitution part.
     
  7. Mar 4, 2013 #6
    Yea I see where I messed up I fixed that above. Could the algebraically inclined show me the manipulation to get the answer?
     
  8. Mar 4, 2013 #7

    SammyS

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    Factor [itex]\displaystyle \ (4-x^2)^{3/2}\ [/itex] from your result & collect terms.
     
  9. Mar 4, 2013 #8
    I think an easier way to do this would be make u=x^2 right from the beginning. Then after that make another substitution y=4-u. Then it's pretty easy from there.
     
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