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Evaluating Integral

  1. Nov 23, 2005 #1
    Hello All

    I am having problem with solving integrals by subsitution.

    I have the following problem. Can anybody help?

    (large S) X^2 /(SQRT(1-X)) DX

    My first and only thought was to subsitute 1 - x with U but that gave me an answer not ever close.

    Does anybody have any tips?


  2. jcsd
  3. Nov 23, 2005 #2


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    Powp, here it is in LaTex:

    [tex]\int \frac{x^2}{\sqrt{1-x}}dx[/tex]

    So let:


    and work it though completely. That is, then what is [itex]u^2[/itex]?

    What then is x in terms of u?

    What then is dx in terms of u and du?

    What is [itex]x^2[/itex] in terms of u?
  4. Nov 23, 2005 #3
    Thanks for the reply

    u^2 = 1 - x
    2u * du = -1 * dx
    -2u * du = dx

    x = 1 - u^2

    but where do I go from here.
  5. Nov 23, 2005 #4


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    Put them into the integral and DO it!
    Your original integral was
    [tex]\int \frac{x^2}{\sqrt{1-x}}dx[/itex]
    If you let [itex]u= \sqrt{1-x}= (1-x)^\frac{1}{2}[/itex] then, yes, [itex]u^2= 1-x[/itex] so 2u du= -dx and x= 1- u2 so x2= (1- u2)2= 1- 2u2+ u4. Your integral becomes
    [tex]\int \frac{1- 2u^2+ u^4}{u}(-2u du)= -2\int \left(1- 2u^2+ u^4\right)du[/itex].
    Actually, your first thought: substituting for 1- x works perfectly well.
    If u= 1-x then du= -dx. Also, x= 1- u so x2= (1- u)2= 1- 2u+ u2. The integral becomes
    [tex]-\int \frac{1- 2u+ u^2}{u^\frac{1}{2}}du= -\int \left(u^{-\frac{1}{2}}- 2u^\frac{1}{2}+ u^\frac{3}{2}\right) du[/tex]
    Last edited: Nov 23, 2005
  6. Nov 23, 2005 #5


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    We're makin' progress. Now you need to substitute all these expressions into the original integral:


    So, substitute each one. I'll do one of them:


    There you go. Substitute -2udu for dx in the integral. Well . . . do the same for [itex]x^2[/itex] and [itex]\sqrt{1-x}[/itex] . . . what does the integral then look like in terms of u? It's a lot easier now right? So solve it for u then don't forget to substitute back [itex]\sqrt{1-x}[/itex] in place of u in the final answer.

    Edit: Didn't see Hall's reply but you know what to do now.
    Last edited: Nov 23, 2005
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