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Homework Help: Definite integral involving partial fractions

  1. Sep 22, 2015 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    trigonometric identities

    3. The attempt at a solution
    I did a trig substitution of u=tan(θ/2) and from that I could substitute cos(θ) = 1-u2/1+u2

    dθ = 2/(1+u2)
    du = 1/2 sec2(θ/2) dθ

    I simplified a bit and changed the bounds to get 2du/(5u2 + 1)(1 + u2)2 with lower bound 0 and upper bound 1.

    I think at this point I have to do partial fractions. Do I need 3 linear terms or 2? I tried Ax+B/5u2+1 + Cx + D/(1+u2)2 + Ex+F/(1+u2) = 2/(5u2 + 1)(1 + u2)

    I have no idea how to evaluate such a complicated system of equations. I could really use some guidance here.
    Last edited: Sep 22, 2015
  2. jcsd
  3. Sep 22, 2015 #2


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    First you need to fix up the errors.

    You have written x in the numerators where you should have written u.
    You have omitted parentheses around your numerators and denominators.
    You have omitted the exponent 2 in the second factor of the denominator on the RHS.

    Once you've done that, multiply both sides by the RHS denominator and you'll have a polynomial equation of order 6 with six unknown constants. Put it all on one side and then you get seven equations by equating the coefficient of each power of u to zero.

    Those equations will enable you to solve to find the constants.
  4. Sep 22, 2015 #3


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    Throw some parentheses in where needed, and that looks right.
    There should be a du on the right hand side of the above.
    ##\displaystyle \ d\theta=\frac{2}{1+u^2}du \ ##

    I don't see any need to use the following.
    Check your Algebra. I get that the (1+ u2) should cancel, not be squared.
    Last edited: Sep 23, 2015
  5. Sep 23, 2015 #4
    thanks a ton. I didn't catch that at first. I actually re-checked my algebra like 10 times and kept getting different answers. It's sad to see gaps in my knowledge but I guess this gives me good reason to strengthen those areas.
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