Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.6 Strategies for Integration - Exercises - Page 431: 40


$$ 8\left( \frac{x}{4}\frac{\sqrt{16-x^2}}{4}- \ln | \frac{x}{4}+\frac{\sqrt{16-x^2}}{4}| \right)+C$$

Work Step by Step

Given $$ \int \sqrt{x^{2}-16} d x$$ Let \begin{align*} x&=4\sec u\\ dx&=4\sec u\tan udu \end{align*} Then \begin{align*} \int \sqrt{x^{2}-16} d x&= 4\int \sqrt{16\sec^2 u-16} \sec u\tan udu\\ &=4\int \sqrt{16\tan^2 u-} \sec u\tan udu\\ &=16\int \sec u\tan^2 udu\\ &=16\int (\sec^3 u-\sec u )du,\ \ \\ & \text{From the table: }\\ &= 16\left(\frac{1}{2} \sec u \tan u+\frac{1}{2} \ln |\sec u+\tan u|-\ln |\sec u+\tan u| \right)+C\\ &= 16\left(\frac{1}{2} \sec u \tan u-\frac{1}{2} \ln |\sec u+\tan u| \right)+C\\ &= 8\left( \frac{x}{4}\frac{\sqrt{16-x^2}}{4}- \ln | \frac{x}{4}+\frac{\sqrt{16-x^2}}{4}| \right)+C \end{align*}
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