## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{\pi}{2}$
Consider $f(x)=\int_{0}^{1} \dfrac{1}{\sqrt{1-x^2}} dx$ suppose $\sin u =x \implies dx=\cos u du$ Thus, $\int_{0}^{a} \dfrac{1}{\sqrt{1-x^2}} dx=\int_{0}^{\sin ^{-1} a} \dfrac{\cos u}{\sqrt{1-\sin^2 u}} dx$ and $\lim\limits_{a \to 1^{-}} \int_{0}^{\sin ^{-1} a} \dfrac{\cos u}{\sqrt{1-\sin^2 u}} dx=\lim\limits_{a \to 1^{-}}\sin^{-1} a$ or, $\lim\limits_{a \to 1^{-}}\sin^{-1} a =\dfrac{\pi}{2}$