Repeaded Integrals again

[Bucky]

Homework Statement

Evaluate the following repeated integral:

$$\int^2_0 dv \int^\frac{1}{2}_0 \frac{v}{\root(1-u^2) du}$$

The Attempt at a Solution

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


$$\int^2_0 dv \int^\frac{1}{2}_0 1 - u^2)^\frac{1}{2} du}$$

integrating using standard integral

$$(ax+b)^n = \frac{(ax+b)^{n+1}}{a(n+1)}$$

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

which i solved to get.

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

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?)

 

[arunbg]

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.