University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.3 - Trigonometric Substitutions - Exercises - Page 439: 3



Work Step by Step

Let us consider $x =2\tan \theta $ and $dx= 2 \sec^2 \theta d \theta $ Now, the given integral becomes: $\int \dfrac{2 \sec^2 \theta d \theta }{\sqrt {4+4 \tan^2 \theta)}} $ or, $=\dfrac{1}{2} \int d \theta$ Now, integrate with limits: $\int_{-2}^{2} \dfrac{1}{2} d \theta=\dfrac{1}{2}[\tan^{-1} (1)-\tan^{-1} (-1)]=\dfrac{\pi}{4}$
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.