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: 33


$\sqrt{1-\sin^2t}-\ln |\dfrac{1+\sqrt{1-\sin^2t}}{\sin t} |+C$

Work Step by Step

Re-write as $I=\int\dfrac{\cos t}{\sin t}\sqrt{1-\sin^2t}dt$ Suppose $u=\sin t \implies du=\cos (t) dt$ Now, $I=\int\dfrac{\sqrt{1-u^2}}{u} \space du \\=\sqrt{1-u^2}-\ln |\dfrac{1+\sqrt{1-u^2}}{u} |+C \\=\sqrt{1-\sin^2t}-\ln |\dfrac{1+\sqrt{1-\sin^2t}}{\sin t} |+C$
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