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Integration which possibly involves partial fractions.

  1. Jan 9, 2014 #1
    1. The problem statement, all variables and given/known data

    Well this is part of an integration process, namely:

    [tex] \int \frac {sin^2x}{4+3cos^2x}dx [/tex]

    2. Relevant equations

    My attempt involved using a u-substitution, namely t = tan x

    3. The attempt at a solution

    Using t = tan x, [tex] sin^2 x = \frac {t^2}{1+t^2} [/tex] and [tex]cos^2 x = \frac {1}{1+t^2}[/tex]

    we have


    [tex] \int \frac {sin^2x}{4+3cos^2x}dx [/tex]

    which becomes

    1. [tex]\int \frac {t^2}{(1+t^2)(4+ \frac {3}{1+t^2})} \times \frac {1}{1+t^2} [/tex]

    2. [tex] \int \frac {t^2}{(1+t^2)( \frac {4+4t^2+3}{1+t^2})} \times \frac {1}{1+t^2} [/tex]

    and finally yields

    [tex] \int \frac {t^2}{(7+4t^2)(1+t^2)}dt [/tex]

    So i am basically stuck at this point and cannot proceed further.
    Maybe i should have used another substitution or I started it wrongly.
    I cannot seem to get to the solution though i suspect arctan might be present in the answer.
    Please do guide me.
    Thanks again.
     
    Last edited: Jan 9, 2014
  2. jcsd
  3. Jan 9, 2014 #2

    SteamKing

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    It's not clear why you chose t = tan x as your substitution.

    It's also not clear how you obtained the integral in t after substituting. You clearly wound up with a more complicated integral using the t substitution. You should show all the steps you took to make sure you haven't made any mistakes.
     
  4. Jan 9, 2014 #3
    What method do u think would be best to tackle this one? well i chose tan x because with the denominator containing terms squared, i thought i would end up with an arctan case, but the [tex] t^2 [/tex] in the numerator after simplifying makes it harder to achieve the form if not impossible.

    Also when t = tan x, [tex] sin^2 x = \frac {t^2}{1+t^2} [/tex] and [tex]cos^2 x = \frac {1}{1+t^2}[/tex]

    Upon substituting including the dx i ended up simplifying it like i showed.

    I am still trying some other methods...
     
    Last edited: Jan 9, 2014
  5. Jan 9, 2014 #4

    haruspex

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    Partial fractions, as you suggested. Did you try that?
     
  6. Jan 9, 2014 #5
    It will be of the form

    [tex] \frac {At+B}{1+t^2} + \frac {Ct+D}{7+4t^2} [/tex] right?

    Consequently

    For the numerator,

    [tex]t^2 = (At+B)(7+4t^2) + (Ct+D)(1+t^2)[/tex]

    [tex] (4A+C)t^3 + (4B+D) t^2 + (7A+C)t + (7B+D) = t^2[/tex]

    Comparing coefficients of t^2 and comparing constants we get the 2 equations respectively;

    4B + D = 1 and 7B+D = 0

    which gives B = - 1/3 and D = 7/3
    A and C being both 0

    Thus the partial fraction is

    [tex] \frac {7}{3(7+4t^2)} - \frac {1}{3(1+t^2)}[/tex]

    But on expanding it does not give t^2 in the numerator, am I doing something wrong?
     
    Last edited: Jan 9, 2014
  7. Jan 9, 2014 #6

    haruspex

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    Since your integrand fraction only has t2 terms, no odd powers, the partial fraction expansion must have A = C = 0. (Imagine substituting u = t2 first.)
     
  8. Jan 9, 2014 #7
    Alright. But it should give the same answer as in my edited post
     
  9. Jan 9, 2014 #8

    haruspex

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    Check that again. It does for me.
     
  10. Jan 9, 2014 #9
    Oh god yes, i forgot the 9 in the denominator. Well that partial fraction makes the integrand now easier.
    I should complete the denominators to the square and use arctan integral?
     
  11. Jan 9, 2014 #10

    haruspex

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    Use arctan, yes. No need to complete any squares.
     
  12. Jan 9, 2014 #11
    Thank you!
     
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