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Trig Substitution

  1. Feb 26, 2013 #1
    1. The problem statement, all variables and given/known data
    evaluate the integral.


    2. Relevant equations
    integral (x^3 / (x^2 - 16)


    3. The attempt at a solution
    x=4sec∅
    dx=4sec∅tan∅d∅

    1. i substituted those values in, and then split sec^4∅ into sec^2∅ and (1+tan^2∅).
    2. integral 16 (1/u) du + integral 16 (u) du.
    3. end with:

    16 * ln|sqrt(x^2 - 16) / 4| + (1/2) (x^2 - 16)


    however, the answer is:

    (x^2)/2 + 8 * ln|x^2 - 16| + c.
     
  2. jcsd
  3. Feb 26, 2013 #2
    okay. i tried u=x^2 -16

    1/2 integral (u+16) / u du

    1/2 integral (du) + 1/2 integral (16/u) du

    1/2(x^2 - 16) + 8 * ln|x^2 - 16|

    i get part of the correct answer: (x^2 / 2) + 8 * ln|x^2 - 16| but i have an extra -8 .

    correct answer: (x^2 / 2) + 8 * ln|x^2 - 16| + c
    my answer: (x^2 / 2) + 8 * ln|x^2 - 16| - 8 + c
     
  4. Feb 26, 2013 #3

    Dick

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    Science Advisor
    Homework Helper

    If you differentiate both of those you get the same thing. They are both correct. You can absorb the -8 into the +c.
     
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