Use the substitution x = sin\theta to find the exact value of;

$\displaystyle \int^{\frac{1}{2}}_0 \frac{1}{( {1-x^{2}})^\frac{3}{2}} $

$\displaystyle \int \frac{1}{(1-sin^{2}\theta)^{\frac{3}{2}}} $

$\displaystyle \frac{dx}{d\theta} = cos\theta $

so when I put this all together I get ;

$\displaystyle \int \frac{1}{(cos^{2}\theta)}^{\frac{3}{2}} cos\theta d\theta $

and there does not see to be much I can do from here, cant seem to cancel the cosines,

and the 3/2 power is suppose to be at the bottom, of the fraction. Cant seem to do it on latex