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U substitution

  1. Nov 22, 2008 #1
    1. The problem statement, all variables and given/known data

    Integrate: x^2 dx / (1-x)^1/2

    2. Relevant equations

    U substitution

    3. The attempt at a solution

    First I defined u = (1-x)^1/2
    du = -dx/2(1-x)^1/2
    dx = -2(1-x)^1/2 du

    then for x
    u^2 = 1-x
    x = 1-u^2

    integral of: (1-u^2)^2/u
    = (u^4-2u^2+1) / u
    = u^3-2u+1/u


    u^4/4 - u^2 + 1/u

    = ((1-x)^2)/4 - (1-x) + ln|1-x| + C

    Is this right?
  2. jcsd
  3. Nov 22, 2008 #2
    You can try differentiating your answer to check if it is correct..
  4. Nov 22, 2008 #3
    I differentiated and the answer I got was wrong. Does anyone know what I did wrong?
  5. Nov 22, 2008 #4
    looks like you should have gotten ,

    ln| (1-x)^1/2 |

    EDIT: I was wrong here, look at Dick's hint later. It seems pretty easy after his hint.
    Last edited: Nov 22, 2008
  6. Nov 22, 2008 #5


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    You figured out that dx=-2u*du. But then you never used it. There's no logs in the problem at all.
  7. Nov 22, 2008 #6
    If dx = -2u*du that means that I would have to set u = (1-x)^1/2 and square both sides and then take the derivative, right? Well, if I do that, what do I do for the x^2 on the numerator?
  8. Nov 22, 2008 #7


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    What?? No, you are doing everything right, but you forgot to replace the dx with an expression for du. You said "dx = -2(1-x)^1/2 du". Use that expression.
  9. Nov 22, 2008 #8
    Ok, I reworked the dx part and I got

    Integral of:
    = -2(1-u^2)^2
    = -2(u^4-2u^2+1)

    = -(2/5)(u^5) + (4/3)(u^3) - 2u + C

    = -(2/5)((1-x)^5) + (4/3)((1-x)^3) - 2(1-x) + C

    But when I differentiate this Im not getting x^2/sqrt(1-x)
  10. Nov 22, 2008 #9


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    u=sqrt(1-x). You substituted u=(1-x).
  11. Nov 22, 2008 #10
    After fixing the substitution

    -(2/5)((sqrt(1-x))^5) + (4/3)((sqrt(1-x))^3) - 2(sqrt(1-x)) + C

    d/dx of -(2/5)((sqrt(1-x))^5) + (4/3)((sqrt(1-x))^3) - 2(sqrt(1-x)) + C

    = 1/sqrt(1-x) - (x+1)*(sqrt(1-x))

    which isnt x^2/(sqrt(1-x))

    I went to mathematica's online integral calculator and they gave this:

    My Ti-89 gave the same answer once factored. I still dont know what I'm doing wrong.
  12. Nov 22, 2008 #11


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    1/sqrt(1-x) - (x+1)*(sqrt(1-x)). Multiply the numerator and denominator of the second term by sqrt(1-x). 1/sqrt(1-x)-(x+1)(1-x)/sqrt(1-x)=(1-(1-x^2))/sqrt(1-x)=x^2/sqrt(1-x). It's right. You've got too many machines telling you what to do.
  13. Nov 22, 2008 #12
    Thank you so much
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