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: 4



Work Step by Step

Let us consider $x =2\tan \theta $ and $dx= 2 \sec^2 \theta d \theta $ Now, the given integral becomes: $\int \dfrac{2 \sec^2 \theta d \theta }{8 \sec^2 \theta} =\dfrac{1}{4} \int d \theta$ Now, integrate with limits: $\int_{0}^{2} \dfrac{1}{4} d \theta=[\dfrac{1}{4} \theta +C]_{0}^{2}=\dfrac{1}{4}[\tan^{-1} (1)-\tan^{-1} (0)]=\dfrac{\pi}{16}$
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