FTC # 2- the d d x [ ∫ a ( x ) b ( x ) F ( x ) d x ] {\displaystyle {\frac {d}{dx}}\left[\int \limits _{a(x)}^{b(x)}F(x)dx\right]} is F(b)-F(a) where F is the antiderivitive of f such that F ′ = f {\displaystyle F^{\prime }=f}
37) g(x)= ∫ 1 / 2 3 / 2 6 1 − t 2 d t {\displaystyle \int \limits _{1/2}^{{\sqrt {3}}/2}{\frac {6}{\sqrt {1-t^{2}}}}dt}
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 {\displaystyle g^{\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}}
= 6 s i n − 1 ( 3 ) / 2 ) − ( 6 s i n − 1 ( 1 / 2 ) ) {\displaystyle 6sin^{-1}({\sqrt {3}})/2)-(6sin^{-1}(1/2))}
= π {\displaystyle \pi }
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![{\displaystyle {\frac {d}{dx}}\left[\int \limits _{a(x)}^{b(x)}F(x)dx\right]}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/9331129b1175ef32e6c2e77b2e98e903c62fa668)
is F(b)-F(a) where F is the antiderivitive of f such that 
