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Integration by substitution

  1. Nov 26, 2007 #1
    the problem asks for the area under the shaded region of the line y = 1/(1-x^2) on the interval [-1,1].

    so far i've set up the integral showing
    \int [tex]dx/(1-x^2)[\tex]
    on the interval [-1,1]


    i'm pretty sure you have to use substitution to solve it, but i cant seem to figure it out
    please show/describe each step to solve it.
     
    Last edited: Nov 26, 2007
  2. jcsd
  3. Nov 26, 2007 #2
    I'll give you a hint. You will start with trig substitution.
     
  4. Nov 26, 2007 #3
    [tex]\int \frac{dx}{1-x^2}=\int \frac{dx}{(1+x)(1-x)}[/tex]

    and do partial fractions, or you can do the trig substitution.
     
  5. Jan 26, 2008 #4
    Well, to not get anyone confused, i'm taking out my calculations, cus its like 5 o clock in teh morning and i don't quite remember all the steps i was doing, so i might re attack this later , when i'm awake.
     
    Last edited: Jan 27, 2008
  6. Jan 26, 2008 #5

    Mute

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    That is incorrect. You made the substitution [itex]u = 1 - x^2[/itex] in the integrand, but failed to make the change [itex]dx \rightarrow -\frac{du}{2x}[/itex] (and then substituting in the expression for x in terms of u, which is multivalued on [-1,1], so we'd want to reduce the integration range to [0,1] by noting the integrand is even).

    Hence, the correct integral to evaluate, using that subsitution, is

    [tex]2\int_1^0 du~\frac{1}{u\sqrt{1-u}}[/tex]

    which really isn't much better.
     
  7. Jan 27, 2008 #6
    Darnit, your right, i was typing that in there after figuring it out, and i knew i did something wrong, but i couldn't figure it out. Thanks a lot.
     
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