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Math Help - Integrating the square root of a polynomial

  1. #1
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    Integrating the square root of a polynomial

    How would I begin to solve this integrand ? by completing the square ?

    I try completing the square when I factor out a - 4 outside the square root but it does not work out quite right.


    \int{\frac{1}{\sqrt{-4x^2+12x-4}}}

    This one is just a tad confusing to me when trying to complete the square, which I would assume would be the best way to do this.
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  2. #2
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    Quote Originally Posted by The Power View Post
    How would I begin to solve this integrand ? by completing the square ?

    I try completing the square when I factor out a - 4 outside the square root but it does not work out quite right.


    \int{\frac{1}{\sqrt{-4x^2+12x-4}}}

    This one is just a tad confusing to me when trying to complete the square, which I would assume would be the best way to do this.
    Note that -4x^2 + 12x - 4 = -4(x^2 - 3x + 1) = -4\left[\left( x - \frac{3}{2}\right)^2 - \frac{5}{4}\right] = 4 \left[\frac{5}{4} - \left( x - \frac{3}{2}\right)^2\right]

     = 5 - (2x - 3)^2 = (\sqrt{5})^2 - (2x - 3)^2 .
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    So, I was going in the general direction correctly by using completing the square method ?
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    Hi,

    I dont really know where fantastic is going with his tip but the problem looks like an arcsin/arccos to me,

    the definition is:

    derivative of arcsin(x) is:



    derivative of arccos(x) is:




    Make it look like 1 + something^2 in the square root and the solution should not be far away.
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    Quote Originally Posted by knivmannen View Post
    Hi,

    I dont really know where fantastic is going with his tip but the problem looks like an arcsin/arccos to me,

    the definition is:

    derivative of arcsin(x) is:



    derivative of arccos(x) is:




    Make it look like 1 + something^2 in the square root and the solution should not be far away.
    I think you meant "...1 - something^2...", and this is what I believe fantastic was heading to: he already got that the expression within the square root is (sqrt(5))^2 - (2x - 3)^2 = 5 {1 -[(2x-3)/Sqrt(5)]^2}, and if we get this into the square root in the denominator this is exactly an Arcsin integral.

    Tonio
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    Quote Originally Posted by mr fantastic View Post
    [tex] = 5 - (2x - 3)^2
    Where that last bit comes from ? I see where 5 comes from, but not
    (2x - 3)^2
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    Quote Originally Posted by The Power View Post
    Where that last bit comes from ? I see where 5 comes from, but not
    (2x - 3)^2


    First, check that 5 - (2x - 3)^2 really equals what it must; second, in fantastic's post, just before he obtained this expression, he took out (1/4) from the parentheses but here he got;

    [5/4 - (x - 3/2)^2] = [5/4 - ((2x - 3)/2)^2] = [5/4 - (1/4)(2x-3)^2] and now 1/4 goes out and etc.

    Tonio
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    Still dont see where 1/4 and 2x magically comes from
    Last edited by The Power; October 7th 2009 at 11:11 AM.
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  9. #9
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    Quote Originally Posted by The Power View Post
    Still dont see where 1/4 and 2x magically comes from
    Ok, this already is high school algebra, so "get your hands dirty" and begin working on it (at the very least grab a pencil and a sheet of paper and write down both fantastic's message and mine to see clearly what's written there...and then begin doing maths!)

    Tonio
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    Ah, I see, he expanded out (x - 3/2) ^2 the multiplied, through -4 so techincally his simplifaction isn't needed just more easier.
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    Quote Originally Posted by The Power View Post
    Ah, I see, he expanded out (x - 3/2) ^2 the multiplied, through -4 so techincally his simplifaction isn't needed just more easier.

    I don't know what he did, but no need at all to expand anything:

    (x - 3/2)^2 = ([2x-3]/2)^2 = (1/4)*(2x-3)^2

    Tonio
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    Do I use the trig. substitution of arcsin x ?
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  13. #13
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    Quote Originally Posted by The Power View Post
    Do I use the trig. substitution of arcsin x ?

    I supsoe you can try substitution, but in fact:

    (1/5)[1/Sqrt(1 - [(2x-3)/Sqrt(5)]^2] =

    (1/2Sqrt(5)))[(2/Sqrt(5))/Sqrt(1 - [(2x-3)/Sqrt(5)]^2] ==>

    the integral of the above equals (1/2Sqrt(5))*Arcsin[(2x-3)/sqrt(5)]

    Tonio
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    Mind throwing some math tags around those ?

    Couldn't I use 2x - 3 = sin theta and solve for x, I tried that but its seems to complicate things
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  15. #15
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    Quote Originally Posted by The Power View Post
    Mind throwing some math tags around those ?

    Couldn't I use 2x - 3 = sin theta and solve for x, I tried that but its seems to complicate things
    tonio has graciously walked you through basic algebra that a student attempting questions like the one you posted should have had no trouble with. So if you want math tags, I suggest you quote what tonio wrote and then insert the tags yourself into the quote. And the least you could do is to actually acknowledge the help you got by saying thankyou (or clicking on the Thanks button).

    If you follow what's been posted in this thread then it should be clear that you have a standard form. I had hoped the basic substitution u = 2x - 3 would have made that obvious. I suggest you go back and review this material.
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