## Calculus, 10th Edition (Anton)

Published by Wiley

# Chapter 7 - Principles Of Integral Evaluation - 7.3 Integrating Trigonometric Functions - Exercises Set 7.3 - Page 507: 28

#### Answer

$$2\ln \left| {\sec \sqrt x + \tan \sqrt x } \right| + C$$

#### Work Step by Step

\eqalign{ & \int {\frac{{\sec \left( {\sqrt x } \right)}}{{\sqrt x }}} dx \cr & = \int {\frac{{\sec {x^{1/2}}}}{{{x^{1/2}}}}} dx \cr & = \int {\sec {x^{1/2}}\left( {{x^{ - 1/2}}} \right)} dx \cr & {\text{substitute }}u = {x^{1/2}},{\text{ }}du = \frac{1}{2}{x^{ - 1/2}}dx \cr & 2du = {x^{ - 1/2}}dx \cr & = \int {\sec {x^{1/2}}\left( {{x^{ - 1/2}}} \right)} dx \cr & = \int {\sec u\left( {2du} \right)} \cr & = 2\int {\sec udu} \cr & {\text{find the antiderivative }}\left( {{\text{use formula 22 page 503}}} \right) \cr & = 2\ln \left| {\sec u + \tan u} \right| + C \cr & {\text{write in terms of }}x,{\text{ replace }}u = {x^{1/2}} \cr & = 2\ln \left| {\sec {x^{1/2}} + \tan {x^{1/2}}} \right| + C \cr & = 2\ln \left| {\sec \sqrt x + \tan \sqrt x } \right| + C \cr}

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