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: 24


$\dfrac{1}{4 \sqrt 3}$

Work Step by Step

Let us consider $x =2 \sin \theta $ and $dx=2 \cos \theta d \theta $ and $2 \cos \theta =\sqrt{4-x^2}$ Now, the given integral becomes (without limits): $\int \dfrac{2 \cos \theta d \theta }{8 \cos ^3 \theta} =\dfrac{1}{4} \int \sec^2 \theta d\theta$ Now, integrate with limits and plug in $\tan \theta =\sin\theta/\cos \theta=\dfrac{x}{\sqrt{4-x^2}}$: $\dfrac{1}{4} \int\tan \theta +C=(\dfrac{1}{4})[\dfrac{x}{(4-x^2)^{1/2}}]_0^{1}=(\dfrac{1}{4})[\dfrac{1}{(4-1)^{1/2}}-0]=\dfrac{1}{4 \sqrt 3}$
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