Answer
$2\sqrt 3$
Work Step by Step
$\int_{0}^{\frac{\sqrt 3}{2}}\frac{1}{(1-t^{2})^\frac{5}{2}}dt$
$let$ $t=sinθ$
$dt=cosθdθ$
$0=sinθ$
$θ=0$
$\frac{\sqrt 3}{2}=sinθ$
$θ=\frac{\pi}{3}$
$\int_{0}^{\frac{\pi}{3}}\frac{1}{(1-sin^{2}θ)^\frac{5}{2}}(cosθ)dθ$
Through the Pythagorean identity $sin^{2}θ+cos^{2}θ=1$, $1-sin^{2}θ=cos^{2}θ$:
$\int_{0}^{\frac{\pi}{3}}\frac{cosθ}{(cos^{2}θ)^\frac{5}{2}}dθ$
$\int_{0}^{\frac{\pi}{3}}\frac{cosθ}{cos^{5}θ}dθ$
$\int_{0}^{\frac{\pi}{3}}\frac{1}{cos^{4}θ}dθ$
$\int_{0}^{\frac{\pi}{3}}sec^{4}θdθ$
$\int_{0}^{\frac{\pi}{3}}sec^{2}θsec^{2}θdθ$
Through the Pythagorean identity $tan^{2}θ+1=sec^{2}θ$:
$\int_{0}^{\frac{\pi}{3}}(tan^{2}θ+1)sec^{2}θdθ$
$let$ $u=tanθ$
$du=sec^{2}θdθ$
$u(0)=0$
$u(\frac{\pi}{3})=\sqrt 3$
$\int_{0}^{\sqrt 3}(u^{2}+1)du$
$[\frac{1}{3}u^{3}+u]_{0}^{\sqrt 3}$
$[\frac{1}{3}(\sqrt 3)^{3}+\sqrt 3-(\frac{1}{3}(0)^{3}+0)]$
$[\sqrt 3+\sqrt 3]$
$2\sqrt 3$
