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: 28



Work Step by Step

Plug $r^2=p \implies dp=2rdr$ Here, we have $\lim\limits_{a \to 1^{-}}\int_{0}^{a} \dfrac{4r dr}{\sqrt {1-r^4}}=\lim\limits_{a \to 1^{-}}\int_{0}^{a} \dfrac{2 dp}{\sqrt {1-u^2}}$ or, $\lim\limits_{a \to 1^{-}}\int_{0}^{a} \dfrac{2 dp}{\sqrt {1-u^2}}=\lim\limits_{a \to 1^{-}}[2\sin^{-1} (p)]_0^a$ This implies that $\lim\limits_{a \to 1^{-}}(2 \sin^{-1} a-2 \sin^{-1} (0))=(2) (\dfrac{\pi}{2})-0$ Thus, $(2) (\dfrac{\pi}{2})=\pi$
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