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Repeaded Integrals again

  1. Mar 22, 2007 #1
    1. The problem statement, all variables and given/known data

    Evaluate the following repeated integral:

    [tex] \int^2_0 dv \int^\frac{1}{2}_0 \frac{v}{\root(1-u^2) du} [/tex]


    3. The attempt at a solution

    since 1/something is something ^(-1), and root something is something^(1/2), i rearragned to get..


    [tex] \int^2_0 dv \int^\frac{1}{2}_0 1 - u^2)^\frac{1}{2} du} [/tex]

    integrating using standard integral

    [tex] (ax+b)^n = \frac{(ax+b)^{n+1}}{a(n+1)} [/tex]

    [tex] \int^2_0 [ \frac{(1 - u^2)^{1/2}}{-1(\frac{v}{2})} [/tex]

    which i solved to get.

    [tex]\int^2_0 -\frac{2(\frac{3}{4})^{3/4}}{v} dv
    [/tex]

    given that the answer in the book is pi over 3, i think i've made a mistake (also that 3/4 ^ 3/4 doesn't look right since surley it won't resolve well?)
     
  2. jcsd
  3. Mar 22, 2007 #2
    The inner integral is not a linear function of the dependent variable ie, it is not of the form (ax+b)^n, it is more like (ax^2+b)^n. In your case, there is a very simple answer for the integral you seek. Check your integration tables or better still, come up with the solution on your own.
     
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