## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 7: Transcendental Functions - Section 7.2 - Natural Logarithms - Exercises 7.2 - Page 381: 48

#### Answer

$$\ln \left| {2 + \sec y} \right| + C$$

#### Work Step by Step

\eqalign{ & \int {\frac{{\sec y\tan y}}{{2 + \sec y}}} dy \cr & {\text{Use substitution:}}\cr & {\text{Let }}u = 2 + \sec y,{\text{ so that }}du = \sec y\tan ydy \cr & {\text{Write the integral in terms of }}u \cr & \int {\frac{{\sec y\tan y}}{{2 + \sec y}}} dy = \int {\frac{{\sec y\tan y}}{u}} \left( {\frac{{dy}}{{\sec y\tan y}}} \right) \cr & = \int {\frac{1}{u}} \left( {\frac{{du}}{1}} \right) \cr & = \int {\frac{1}{u}} du \cr & {\text{Integrate }} \cr & = \ln \left| u \right| + C \cr & {\text{Write in terms of }}y:\cr & {\text{Replace }}2 + \sec y\tan t{\text{ for }}u \cr & = \ln \left| {2 + \sec y} \right| + C \cr}

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