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: 32

Answer

$\sqrt{x(2-x)}+2\sin^{-1} (\sqrt{\dfrac{x}{2}})+C$

Work Step by Step

Suppose $u=\sqrt x \implies du=\dfrac{1}{2\sqrt x}dx \implies \dfrac{1}{\sqrt{x}}dx=2 \space du$ and Let $x=u^2$ Therefore, $I=2\int\sqrt{2-u^2}du\\=2(\dfrac{u}{2}\sqrt{2-u^2}+\dfrac{2}{2}\sin^{-1}\dfrac{u}{\sqrt2} )+C\\=u\sqrt{2-u^2}+2\sin^{-1}\dfrac{u}{\sqrt2}+C \\=\sqrt x\sqrt{2-x}+2\sin^{-1} (\dfrac{\sqrt x}{\sqrt2}) +C \\=\sqrt{x(2-x)}+2\sin^{-1} (\sqrt{\dfrac{x}{2}})+C$
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.