Integration by substitution help

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

Here... unfortunately, though, you may have to wait for the latex to work.

Did you forget that (x^m)^(n/p) = x^{(mn)/p}, i.e. {(cosx)^2}^(3/2) = (cosx)^{(3*2)/2} = cos^3{x}? (Giggle) Full solution here, but you may have to wait for the latex to work. (Thinking)

We have $\displaystyle \int_{0}^{\frac{1}{2}}\dfrac{1}{\left(1-x^2\right)^{\frac{3}{2}}}\;{dx}$ If we let (as suggested) $\displaystyle x = \sin{\theta}$, then $\displaystyle \dfrac{dx}{d\theta} = \cos{\theta} \Rightarrow dx = d\theta\cos{\theta}.$ When $\displaystyle x = \dfrac{1}{2}$, $\displaystyle \sin\theta = \dfrac{1}{2} \Rightarrow \theta = \dfrac{\pi}{6}$; similarly, when $\displaystyle x = 0$, $\displaystyle \sin{\theta} = 0 \Rightarrow \theta = 0$. Then we have $\displaystyle \int_{0}^{\frac{1}{2}}\dfrac{1}{\left(1-x^2\right)^{\frac{3}{2}}}\;{dx} = \int_{0}^{\frac{\pi}{6}}\dfrac{\cos{\theta}}{\left (1-\sin^2{x}\right)^{\frac{3}{2}}}\;{d\theta}$ $\displaystyle = \int_{0}^{\frac{\pi}{6}}\dfrac{\cos{\theta}}{\left (\cos^2{\theta}\right)^{\frac{3}{2}}}\;{d\theta} = \int_{0}^{\frac{\pi}{6}}\dfrac{\cos{\theta}}{\left (\cos^3{\theta}\right)}\;{d\theta} $ $\displaystyle = \int_{0}^{\frac{\pi}{6}}\dfrac{1}{\left(\cos^2{\th eta}\right)}\;{d\theta} = \int_{0}^{\frac{\pi}{6}}\sec^2{x}\;{d\theta} = \bigg[\tan{x}\bigg]_{0}^{\frac{\pi}{6}} = \tan\dfrac{\pi}{6} = \dfrac{1}{\sqrt{3}}$