Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.2 Trigonometric Integrals - 7.2 Exercises - Page 525: 47

Answer

\[\frac{1}{2}\sin 2x+C\]

Work Step by Step

Let \[I=\int\frac{1-\tan^2 x}{\sec^2 x}dx\] \[I=\int\frac{1-\frac{\sin^2 x}{\cos^2 x}}{\frac{1}{\cos^2 x}}dx\] \[I=\int\left[\cos^2 x-\sin^2 x\right]dx\] \[\left[\cos^2 x-\sin^2 x=\cos 2x\right]\] \[I=\int\cos 2x\:dx=\frac{1}{2}\sin 2x+C\] Hence $I=\frac{1}{2}\sin 2x+C$
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