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: 23


$$\frac{1}{5} \ln \left|\frac{5 x+\sqrt{25 x^{2}+2}}{\sqrt{2}}\right|+C$$

Work Step by Step

Given $$\int \frac{d x}{\sqrt{25 x^{2}+2}} $$ Let $$x=\frac{\sqrt{2}}{5} \tan \theta, \quad d x=\frac{\sqrt{2}}{5} \sec ^{2} \theta d \theta $$ Then \begin{aligned} \int \frac{d x}{\sqrt{25 x^{2}+2}} &=\int \frac{\frac{\sqrt{2}}{5} \sec ^{2} \theta d \theta}{\sqrt{25\left(\frac{\sqrt{2}}{5} \tan \theta\right)^{2}+2}} \\ &=\int \frac{\frac{\sqrt{2}}{5} \sec ^{2} \theta d \theta}{\sqrt{25 \cdot \frac{2}{28} \tan ^{2} \theta+2}} \\ &=\int \frac{\frac{\sqrt{2}}{5} \sec ^{2} \theta d \theta}{\sqrt{2\left(\tan ^{2} \theta+1\right)}} \\ &=\int \frac{\frac{\sqrt{2}}{5} \sec ^{2} \theta d \theta}{\sqrt{2} \sqrt{\sec ^{2} \theta}}\\ &=\frac{1}{5} \int \sec \theta d \theta\\ &=\frac{1}{5} \ln |\sec \theta+\tan \theta|+C\\ &=\frac{1}{5} \ln \left|\frac{5 x+\sqrt{25 x^{2}+2}}{\sqrt{2}}\right|+C \end{aligned} (Note that the $\sqrt 2$ can be combined into the constant C as well.)
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.