Answer
$2\sqrt {3} $
Work Step by Step
Suppose $\sin u=y\implies dy=\cos udu$
Now the limits become:
$y=0$, $u=0$ and for $y=\sqrt3/2$, $u=\pi/3$
Let $I=\int^{\pi/3}_0\dfrac{\cos udu}{(1-\sin^2u)^{5/2}} \\=\int^{\pi/3}_0\dfrac{\cos udu}{(\cos^2u)^{5/2}} \\=\int^{\pi/3}_0\sec^4udu \\=[\dfrac{\sec^2u\tan u}{3}\Big]^{\pi/3}_0+(2/3) \space
\times \int^{\pi/3}_0\sec^2(u) du\\=(1/3) \times (4\sqrt3)+(2/3) \times(\sqrt3-0) \\=\dfrac{4\sqrt3}{3}+\dfrac{2\sqrt3}{3} \\=2\sqrt {3} $
