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Homework Help: Integral with trig substitution

  1. Oct 31, 2006 #1
    The bit of the problem that I'm working on:
    [tex]6\int\frac{dx}{x^2-x+1}[/tex]

    My work:
    [tex]=6\int\frac{dx}{(x^2-x+\frac{1}{4})+1-\frac{3}{4}}[/tex]
    [tex]=6\int\frac{dx}{(x-\frac{1}{2})^2+\sqrt{\frac{3}{4}}^2}[/tex]

    let [tex]x-\frac{1}{2}=\sqrt{\frac{3}{4}}\tan\theta[/tex]
    so [tex]dx=\sqrt{\frac{3}{4}}\sec^2\theta d\theta[/tex]

    [tex]=6\int\frac{\sqrt{\frac{3}{4}}sec^2\theta d\theta}{\frac{3}{4}+\frac{3}{4}tan^2\theta}[/tex]
    [tex]=(6)(\frac{\sqrt{3}}{2})(\frac{4}{3})\int\frac{sec^2\theta d\theta}{1+tan^2\theta}[/tex]
    [tex]=4\sqrt{3}\int d\theta[/tex]
    [tex]=4\sqrt{3}\theta[/tex]

    My answer:
    [tex]=4\sqrt{3}\arctan{\frac{2x-1}{\sqrt{3}}}[/tex]

    An integrator's answer:
    [tex]\frac{2}{\sqrt{3}}\arctan{\frac{2x-1}{\sqrt{3}}}[/tex]

    I don't think the two answers differ by a constant, but I can't find my error.
     
  2. jcsd
  3. Oct 31, 2006 #2

    AKG

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    Homework Helper

    Your answer is 6 times greater than the integrator's. I assume you just didn't include the 6 when you put your integrand into the integrator since you can't write stuff before the integral when using that program. Try using the integrator again, bringing the 6 into the numerator of the integrand.
     
  4. Oct 31, 2006 #3
    :redface:
    Thank you.
     
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