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: 22


$\dfrac{1}{3} (x^2-4)^{3/2}+C$

Work Step by Step

Let us consider $u=x^2-4 \implies du =2x dx$ Now, the given integral can be written as: $=\dfrac{1}{2} \int 2x \sqrt{x^2-4} dx$ and $=\dfrac{1}{2} \int \sqrt u du$ Now, plug in $u=x^2-4$ Thus, we have $\dfrac{1}{2} \int \sqrt u du=\dfrac{1}{3} (x^2-4)^{3/2}+C$
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