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Partial Fractions (with trig functions)

  1. Oct 24, 2009 #1
    1. The problem statement, all variables and given/known data
    Integral(sinx(x)dx/(cos^2(x)+cos(x)-2)


    2. Relevant equations



    3. The attempt at a solution
    What I tried to do first was factor the denominator, so i got (cos(x)-1)(cos(x)+2)
    from there, I set up my partial fractions equation trying to solve B(cos(x)-1) + A(cos(x)+2) = sin(x)

    at this point i tried plugging in 0 for x -> A(0) + B(3) = 0 ? i didn't feel good about this, after working through the remainder using this i solved for A=0 and B=(-1), which left me with
    integral( -dx/(cos(x)+2), at this point I thought there may have been a 1/2 / double angle formula that I could use to try to simplify the equation; cosx^2 = 1+cos2x / 2, however I havn't felt good about the solving for A=0 and losing a term? any insight would be great, thanks
     
  2. jcsd
  3. Oct 24, 2009 #2
    As far as I know, partial fractions only work with polynomials in the numerator and denominator.
    Try a u-substitution with u=cosx and then partial fractions.
     
  4. Oct 25, 2009 #3

    lurflurf

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    Homework Helper

    Partial fractions work perfectly here. A substitution is not needed to apply partial fractions, here after partial fractions is a good time to change variable because different changes can be used for each term.
    A(0) + B(3) = 0 this is good, but there is a mistake elsewhere as A=0 B=-1 does not work you need another equation.
    I would have started with the identity
    [tex]\frac{\sin(x)}{\cos^2(x)+\cos(x)-2}=\frac{-\cos\left(\frac{x}{2}\right)}{3\sin\left(\frac{x}{2}\right)-2\sin^3\left(\frac{x}{2}\right)}[/tex]
     
    Last edited: Oct 25, 2009
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