## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 8: Techniques of Integration - Section 8.4 - Trigonometric Substitutions - Exercises 8.4 - Page 467: 35

#### Answer

$$\ln \left( {\frac{9}{{\sqrt {10} + 1}}} \right)$$

#### Work Step by Step

\eqalign{ & \int_0^{\ln 4} {\frac{{{e^t}dt}}{{\sqrt {{e^{2t}} + 9} }}} \cr & {\text{We set}}{\text{, }}{e^t} = 3\tan \theta,{\text{ }}{e^t}dt = 3{\sec ^2}\theta d\theta \cr & \theta = {\tan ^{ - 1}}\left( {\frac{{{e^t}}}{3}} \right) \cr & {\text{The new limits on }}\theta {\text{ are found as follows}} \cr & \,\,\,\,\,\,{\text{If }}t = \ln 4,{\text{ then }}\theta = {\tan ^{ - 1}}\left( {\frac{{{e^{\ln 4}}}}{3}} \right) = {\tan ^{ - 1}}\left( {\frac{4}{3}} \right) \cr & \,\,\,\,\,\,{\text{If }}t = 0,{\text{ then }}\theta = {\tan ^{ - 1}}\left( {\frac{{{e^0}}}{3}} \right) = {\tan ^{ - 1}}\left( {\frac{1}{3}} \right) \cr & {\text{With these substitutions}}{\text{, we have}} \cr & \int_0^{\ln 4} {\frac{{{e^t}dt}}{{\sqrt {{e^{2t}} + 9} }}} = \int_{{{\tan }^{ - 1}}\left( {\frac{1}{3}} \right)}^{{{\tan }^{ - 1}}\left( {\frac{4}{3}} \right)} {\frac{{3{{\sec }^2}\theta d\theta }}{{\sqrt {9{{\tan }^2}\theta + 9} }}} \cr & = \frac{1}{3}\int_{{{\tan }^{ - 1}}\left( {\frac{1}{3}} \right)}^{{{\tan }^{ - 1}}\left( {\frac{4}{3}} \right)} {\frac{{3{{\sec }^2}\theta d\theta }}{{\sqrt {{{\tan }^2}\theta + 1} }}} \cr & {\text{use the fundamental identity }}{\tan ^2}x + 1 = {\sec ^2}x \cr & = \int_{{{\tan }^{ - 1}}\left( {\frac{1}{3}} \right)}^{{{\tan }^{ - 1}}\left( {\frac{4}{3}} \right)} {\frac{{{{\sec }^2}\theta d\theta }}{{\sqrt {{{\sec }^2}\theta } }}} \cr & = \int_{{{\tan }^{ - 1}}\left( {\frac{1}{3}} \right)}^{{{\tan }^{ - 1}}\left( {\frac{4}{3}} \right)} {\frac{{{{\sec }^2}\theta d\theta }}{{\sec \theta }}} \cr & = \int_{{{\tan }^{ - 1}}\left( {\frac{1}{3}} \right)}^{{{\tan }^{ - 1}}\left( {\frac{4}{3}} \right)} {\sec \theta d\theta } \cr & {\text{integrating}} \cr & = \left( {\ln \left| {\sec \theta + \tan \theta } \right|} \right)_{{{\tan }^{ - 1}}\left( {1/3} \right)}^{{{\tan }^{ - 1}}\left( {4/3} \right)} \cr & = \left( {\ln \left| {\sec \left( {{{\tan }^{ - 1}}\left( {\frac{4}{3}} \right)} \right) + \tan \left( {{{\tan }^{ - 1}}\left( {\frac{4}{3}} \right)} \right)} \right|} \right) - \left( {\ln \left| {\sec \left( {{{\tan }^{ - 1}}\left( {\frac{1}{3}} \right)} \right) + \tan \left( {{{\tan }^{ - 1}}\left( {\frac{1}{3}} \right)} \right)} \right|} \right) \cr & = \left( {\ln \left| {\frac{5}{3} + \frac{4}{3}} \right|} \right) - \left( {\ln \left| {\frac{{\sqrt {10} }}{3} + \frac{1}{3}} \right|} \right) \cr & = \ln \left| 3 \right| - \ln \left| {\frac{{\sqrt {10} + 1}}{3}} \right| \cr & = \ln \left( {\frac{3}{{\left( {\sqrt {10} + 1} \right)/3}}} \right) \cr & = \ln \left( {\frac{9}{{\sqrt {10} + 1}}} \right) \cr}

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