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Math Help - Integral :Complete the square and give a substitution (not necessarily trigonometric)

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    Integral :Complete the square and give a substitution (not necessarily trigonometric)

    I have tried multiple substitutions and cannot get this homework problem right xD

    \int\dfrac{(x-1)}{\sqrt(2x-x^2)}dx

    everything in the denominator is under the square root
    Last edited by highc1157; September 30th 2010 at 07:04 PM.
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    Quote Originally Posted by highc1157 View Post
    I have tried multiple substitutions and cannot get this homework problem right xD

    \int\dfrac{(x-1)}{\sqrt(2x-x^2)}dx

    everything in the denominator is under the square root
    \displaystyle \int \frac{x-1}{\sqrt{1 -(x^2-2x+1)}} \, dx

    \displaystyle \int \frac{x-1}{\sqrt{1 -(x-1)^2}} \, dx

    can you finish?
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    After completing the square in the squareroot sing, the problem becomes

    \int\frac{1-x}{\sqrt{1 - (x-1)^2}}dx

    Substitue (x-1)^2 = t and proceed.
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    I did that substitution and got (x-2)(x-1), but not same as answer in the book.

    I screwed up somewhere
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    hmmm the book answer is : 1-(1-x)^2

    I tried it again, and got : \sqrt{1-(1-x)^2

    I dont know where i messed up, simplified out everything and this is the integral that I take
    1/2\int{(1-U)^(-(1/2))} du cant fix the exponent, supposed to be (1-U)^(-1/2)
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    \displaystyle \int \frac{x-1}{\sqrt{1 -(x-1)^2}} \, dx

    let t = (x-1)^2

    dt = 2(x-1) \, dx

    \displaystyle \frac{1}{2} \int \frac{2(x-1)}{\sqrt{1 - (x-1)^2}} \, dx

    \displaystyle \frac{1}{2} \int \frac{dt}{\sqrt{1 - t}}

    \displaystyle -\int \frac{-1}{2\sqrt{1 - t}} \, dt

    \displaystyle -\sqrt{1-t} + C

    \displaystyle -\sqrt{1-(x-1)^2} + C

    \displaystyle -\sqrt{2x - x^2} + C
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    that is not the answer in the book , i think i posted the book answer above
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    Quote Originally Posted by highc1157 View Post
    that is not the answer in the book , i think i posted the book answer above
    the "book" answer is incorrect ... they left off the radical and the constant of integration.

    take the derivative of my solution ... what do you get?
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    Hey Skeeter,

    Thanks for you're time and help you put into this , very much appreciated

    But as for your answer and my answer they are slightly different, but this isn't right, what I got? \sqrt{1-(1-x)^2

    oh wow you are right, The derivative of my solution does not equal the original equation !! grrrr lol
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  10. #10
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    Quote Originally Posted by highc1157 View Post
    Hey Skeeter,

    Thanks for you're time and help you put into this , very much appreciated

    But as for your answer and my answer they are slightly different, but this isn't right, what I got? \sqrt{1-(1-x)^2

    oh wow you are right, The derivative of my solution does not equal the original equation !! grrrr lol
    your solution is correct ... note that \sqrt{1-(x-1)^2} = \sqrt{1 - (1-x)^2}
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