Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - 8.3 Trigonometric Substitution - Exercises - Page 410: 27


$$\frac{49}{12 \sqrt{6}}\left[\frac{\sqrt{6 x^{2}-49}}{7} \cdot \frac{\sqrt{6} x}{7}+\ln \left|\frac{\sqrt{6} x}{7}+\frac{\sqrt{6 x^{2}-49}}{7}\right|\right]+C$$

Work Step by Step

Given $$\int \frac{x^{2} d x}{\left(6 x^{2}-49\right)^{1 / 2}}$$ Let $$\sqrt{6}x=7\sec u\ \ \ \ \ \ \to \ \ \ \ \ \sqrt{6}dx=7\sec u\tan udu $$ Then \begin{align*} \int \frac{x^{2} d x}{\left(6 x^{2}-49\right)^{1 / 2}}&=\frac{7}{\sqrt{6}}\frac{49}{6}\int \frac{\sec^3 u\tan udu}{\left(49\sec^2u-49\right)^{1 / 2}}\\ &=\frac{49}{6\sqrt{6}}\int \frac{\sec^3 u\tan udu}{7\tan u}\\ &= \frac{49}{6\sqrt{6}}\int \sec^3 udu \end{align*} Use $$\int \sec ^{n} u d u=\frac{1}{n-1} \tan u \sec ^{n-2} u+\frac{n-2}{n-1} \int \sec ^{n-2} u d u$$ Then $$\int \sec ^{3} u d u=\frac{1}{2} \sec u \tan u+\frac{1}{2} \ln |\sec u+\tan u|+C $$ Hence \begin{align*} \int \frac{x^{2} d x}{\left(6 x^{2}-49\right)^{1 / 2}} &= \frac{49}{6\sqrt{6}}\int \sec^3 udu\\ &= \frac{49}{6\sqrt{6}}\left(\frac{1}{2} \sec u \tan u+\frac{1}{2} \ln |\sec u+\tan u|+C\right) \\ &=\frac{49}{12 \sqrt{6}}\left[\frac{\sqrt{6 x^{2}-49}}{7} \cdot \frac{\sqrt{6} x}{7}+\ln \left|\frac{\sqrt{6} x}{7}+\frac{\sqrt{6 x^{2}-49}}{7}\right|\right]+C \end{align*}
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