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Math Help - stuck on integration

  1. #1
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    stuck on integration

    [(x^2)/(squareroot2x - x^2) ]dx

    do I do u-substitution?
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  2. #2
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    \int\frac{x^{2}}{\sqrt{2x-x^{2}}}dx

    You could let x=u+1, \;\ dx=du

    Make the subs and get:

    \int\frac{(u+1)^{2}}{\sqrt{1-u^{2}}}du

    Now, we can let u=sin(t), \;\ du=cos(t)dt

    Make the subs and it whittles down to:

    \int{(1+sin(t))^{2}}dt

    \int{sin^{2}(t)}dt+2\int{sin(t)}dt+\int{dt}

    \frac{-1}{2}sin(t)cos(t)-2cos(t)+\frac{3}{2}t

    Resub for t=sin^{-1}(u)

    That gives us:

    \frac{3sin^{-1}(u)}{2}-\frac{u\sqrt{1-u^{2}}}{2}-2\sqrt{1-u^{2}}

    Resub for u=x-1 and get:

    \frac{3}{2}sin^{-1}(x-1)-\frac{x\sqrt{2x-x^{2}}}{2}-\frac{3}{2}\sqrt{2x-x^{2}}
    Last edited by galactus; June 6th 2008 at 06:12 PM. Reason: Soory, didn't notice the square root.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by khuezy View Post
    [(x^2)/(squareroot2x - x^2) ]dx

    do I do u-substitution?
    Ok
    we have \int\frac{x^2}{\sqrt{2x-x^2}}dx

    Completing the square we get

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

    Now letting x-1=\sin(\theta)

    so x=\sin(\theta)-1\Rightarrow{dx=\cos(\theta)}

    Giving us

    \int\frac{(\sin(\theta)-1)^2}{\sqrt{1-\sin^2(\theta)}}d\theta

    Can you go from there?
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  4. #4
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    thanks

    I got the integral of (sin(t) +1)^2
    then that becomes sin^2(t) + 2sin(t) + 1

    the integral of sin^2(t)
    is 1/2[1+cos(2t)
    how do I make that in terms of 't'?
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  5. #5
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    How did you get to there?

    You coul
    How did you get to there?

    \frac{-1}{2}sin(t)cos(t)-2cos(t)+\frac{3}{2}t

    how does sin^2(t) go to 1/2sintcost?
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