5.3 The Fundamental Theorem of Calculus/37

$g(x)=\int _{1/2}^{{\sqrt {3}}/2}{\frac {6}{\sqrt {1-t^{2}}}}dtg^{\prime }(x)={\frac {d}{dx}}\left(\int \limits _{1/2}^{{\sqrt {3}}/2}{\frac {6}{\sqrt {1-t^{2}}}}dt\right)=6sin^{-1}(x){\bigg |}_{1/2}^{{\sqrt {3}}/2}=6sin^{-1}({\sqrt {3}})/2)-(6sin^{-1}(1/2))=\pi$