g ( x ) = ∫ 1 2 3 2 6 1 − t 2 d t g ′ ( x ) = d d x ( ∫ 1 / 2 3 / 2 6 1 − t 2 d t ) = 6 s i n − 1 ( x ) | 1 / 2 3 / 2 = 6 s i n − 1 ( 3 ) / 2 ) − ( 6 s i n − 1 ( 1 / 2 ) ) = π {\displaystyle g(x)=\int _{\frac {1}{2}}^{\frac {\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 }
