Answer
$\sqrt 3-\frac{\pi}{3}$
Work Step by Step
$\int_{0}^{\frac{\sqrt 3}{2}}\frac{t^{2}}{(1-t^{2})^\frac{3}{2}}dt$
$let$ $t=sinθ$
$dt=cosθdθ$
$0=sinθ$
$θ=0$
$\frac{\sqrt 3}{2}=sinθ$
$θ=\frac{\pi}{3}$
$\int_{0}^{\frac{\pi}{3}}\frac{sin^{2}θ}{(1-sin^{2}θ)^\frac{3}{2}}(cosθ)dθ$
Through the Pythagorean identity $sin^{2}θ+cos^{2}θ=1$, $1-sin^{2}θ=cos^{2}θ$:
$\int_{0}^{\frac{\pi}{3}}\frac{sin^{2}θcosθ}{(cos^{2}θ)^\frac{3}{2}}dθ$
$\int_{0}^{\frac{\pi}{3}}\frac{sin^{2}θcosθ}{cos^{3}θ}dθ$
$\int_{0}^{\frac{\pi}{3}}\frac{sin^{2}θ}{cos^{2}θ}dθ$
$\int_{0}^{\frac{\pi}{3}}tan^{2}θdθ$
Through the Pythagorean identity $tan^{2}θ+1=sec^{2}θ$, $tan^{2}θ=sec^{2}θ-1$
$\int_{0}^{\frac{\pi}{3}}(sec^{2}θ-1)dθ$
$[tanθ-θ]_{0}^{\frac{\pi}{3}}$
$[tan(\frac{\pi}{3})-\frac{\pi}{3}-(tan0-0)]$
$\sqrt 3-\frac{\pi}{3}$
