Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 8: Techniques of Integration - Section 8.6 - Integral Tables and Computer Algebra Systems - Exercises 8.6 - Page 481: 31



Work Step by Step

Suppose $u=\sqrt {1-x} \implies \dfrac{1}{\sqrt{1-x}}dx=-2\space du$ and $1-x=u^2 \implies x=1-u^2$ Therefore, $I=-2 \int\sqrt{1-u^2} \space du$ or, $I=-2(\dfrac{u}{2}\sqrt{1-u^2}+\dfrac{1}{2}\sin^{-1}u)+C \\=-u\sqrt{1-u^2}-\sin^{-1} (u)+C \\=-\sqrt{x(1-x)}-\sin^{-1}\sqrt{1-x}+C$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.