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Math Help - integrating an expression

  1. #1
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    integrating an expression

    hello again,

    I am trying to integrate this expression and I can't seem to be able to do so. I would greatly appreciate it if you can help me. The expression is: 1/(1 + sqr(2x - x^2)).

    I tried multiplying the denominator with 1- sqr(2x - x^2) but to no avail. I then tried to rewrite the numerator as 1 - x + x, still stuck. I also tried to rewrite the denominator as (u - 2)/2, but I failed to find a solution. anyone any hint, please?
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  2. #2
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    Re: integrating an expression

    Why do you believe you should be able to do it?

    Have you noticed the symmetry about x = 1? And the limited Domain, 0 <= x <= 2?
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  3. #3
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    Re: integrating an expression

    Quote Originally Posted by bibiki View Post
    hello again,

    I am trying to integrate this expression and I can't seem to be able to do so. I would greatly appreciate it if you can help me. The expression is: 1/(1 + sqr(2x - x^2)).

    I tried multiplying the denominator with 1- sqr(2x - x^2) but to no avail. I then tried to rewrite the numerator as 1 - x + x, still stuck. I also tried to rewrite the denominator as (u - 2)/2, but I failed to find a solution. anyone any hint, please?
    looks like a messy one ...

    \frac{1}{1+\sqrt{2x-x^2}} =

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

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

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

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

    I believe the last term can be integrated using a trig substitution ...
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    Re: integrating an expression

    (Sqrt(-(x-2) * x) - (x-1) * sin^(-1) * (1 - x) - 1) / (x-1), where sin^(-1) is an inverse sin function.
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    Re: integrating an expression

    Quote Originally Posted by Will57 View Post
    (Sqrt(-(x-2) * x) - (x-1) * sin^(-1) * (1 - x) - 1) / (x-1), where sin^(-1) is an inverse sin function.
    Wolfram Alpha's solution ...

    integral of 1&#47;&#40;1&#43;sqrt&#40;2x-x&#94;2&#41;&#41; dx - Wolfram|Alpha
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  6. #6
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    Re: integrating an expression

    You WILL need to worry about Domain issues. x = 1 seems to be a problem unless you can prove it isn't.
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    Re: integrating an expression

    TKHunny,
    I believe I should be able to evaluate the integral because it was on a test in the past at the school I go.

    skeeter,
    that is the expression I arrived at, but did not have the idea of reforming it into a trig expression. I guess I will write x - 1 = sin (or perhaps cos), and see what I get.

    Thank you all for your reply.
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  8. #8
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    Re: integrating an expression

    Quote Originally Posted by bibiki View Post
    TKHunny,
    I believe I should be able to evaluate the integral because it was on a test in the past at the school I go.
    No good. It is or it isn't. As soon as the denominator was "rationalized", the Domain issue was introduced unless it can be show that there is no Domain issue. You might impress your teacher if you take a look at it.

    You must answer this question: Are the first and last expressions equivalent? \frac{a}{b}=\frac{a}{b}\cdot\frac{0}{0}=\frac{a \cdot 0}{b \cdot 0}=\frac{0}{0}. If they are, then you're done. If they are not, what are you going to do about the transformation?
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  9. #9
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    Re: integrating an expression

    TKHunny, impress my teacher? I'd rather not before I get into wining and complaining about all the wrong at school, I'll just ask you or skeeter whether I integrated correctly 'the second' expression that skeeter derived. otherwise, I understand you perfectly well, TKHunny. You do have a point.

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

    The substitutions I did are as follows:
    (x - 1) = \sin(t), \mathrm{d}x = \cos(t)\mathrm{d}t
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