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Repeaded Integrals again

  • Thread starter Bucky
  • Start date
  • #1
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Homework Statement



Evaluate the following repeated integral:

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


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?)
 

Answers and Replies

  • #2
590
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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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