University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.7 - Improper Integrals - Exercises - Page 471: 7

Answer

$\dfrac{\pi}{2}$

Work Step by Step

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, $\lim\limits_{a \to 1^{-}}\sin^{-1} a =\dfrac{\pi}{2}$
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