## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 8 - Techniques of Integration - 8.3 Trigonometric Substitution - Exercises - Page 410: 23

#### Answer

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

#### 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.)

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