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More help in integrating

  1. Sep 17, 2007 #1

    rock.freak667

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    Homework Helper

    [SOLVED] More help in integrating

    1. The problem statement, all variables and given/known data

    Find [tex]a,b,n[/tex] such that [tex]a,b[/tex] are rational and n is not an element of Z. a>0,b<0

    [tex]\int_{0}^{1} \frac{3x}{\sqrt{1+6x-3x^2}} = \frac{\pi}{a^n} + b[/tex]

    2. Relevant equations
    [tex]\int \frac{1}{\sqrt{A^2-X^2}} dx= \arcsin{\frac{X}{A}+K[/tex]

    [tex]\int \frac{x}{R} = \frac{R}{a} -\frac{b}{2a}\int\frac{dx}{R} where R=\sqrt{ax^2+bx+c[/tex](Note: this really can't be used as working must be shown)

    3. The attempt at a solution

    [tex]\int_{0}^{1} \frac{3x}{\sqrt{1+6x-3x^2}} dx = \int_0^{1} \frac{3x}{\sqrt{4-3(x-1)^2} }dx[/tex]

    [tex]= \int_{0}^{1} \frac{3x}{\sqrt{(2)^2-(\sqrt{3(x-1)}})^2} dx[/tex]

    Now I can't really see how to get rid of the "x" in the numerator...and way i can get rid of it?(by a nice substitution?
     
    Last edited: Sep 17, 2007
  2. jcsd
  3. Sep 17, 2007 #2

    rock.freak667

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    Homework Helper

    Well a bit more progress has been made:

    Let [tex]\sqrt{3}(x-1)=2sin\theta[/tex]
    [tex]\sqrt{3} dx = \frac{2cos\theta}{\sqrt{3}}[/tex]

    Making the entire integral become:

    [tex]\int\frac{2\sqrt{3}+3}{\sqrt{4-4sin^2\theta}} \frac{2cos\theta}{\sqrt{3}} d\theta[/tex]
    simplifying to give

    [tex]2\int sin\theta d\theta +\int\sqrt{3} d\theta[/tex]

    [tex]= -2cos\theta + \sqrt{3}\theta[/tex]

    [tex]= \left[-2cos({sin^{-1}(\frac{\sqrt{3}(x-1)}{2})) + \sqrt{3}sin^{-1}(\frac{\sqrt{3}(x-1)}{2}\right]_{0}^{1}[/tex]

    giving finally
    [tex] -2+\frac{\pi}{2\sqrt{3}}[/tex] which is not the form I need it in
     
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