Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.5 Partial Fractions - 7.5 Exercises: 69

Answer

\[ = \frac{1}{4}\ln \left| {\frac{{1 - \sin t}}{{1 + \sin t}}} \right| - \frac{1}{2}\frac{1}{{\sin t + 1}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{\sec t}}{{1 + \sin t}}\,dt} \hfill \\ \hfill \\ Given\,\,funcion \hfill \\ \hfill \\ \frac{{\sec t}}{{1 + \sin t}} \hfill \\ \hfill \\ {\text{using}}\,\,{\text{trigonometric identities}} \hfill \\ \hfill \\ \frac{{\sec t}}{{1 + \sin t}} = \frac{1}{{\cos t\,\left( {1 + \sin t} \right)}} \hfill \\ \hfill \\ multiply\,\,the\,\,numerator\,\,and\,\,{\text{denominator}}\,\,by\,\,\cos \,\,t \hfill \\ \hfill \\ = \frac{{\cos t}}{{{{\cos }^2}t\,\left( {1 + \sin t} \right)}} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ = \frac{{\cos t}}{{\,\left( {1 - \sin t} \right)\,{{\left( {1 + \sin t} \right)}^2}}} \hfill \\ \hfill \\ \int_{}^{} {\frac{{\sec t}}{{1 + \sin t}}\,dt} \, = \int_{}^{} {\,\left( {\frac{{\cos t}}{{\,\left( {1 - \sin t} \right)\,{{\left( {1 + \sin t} \right)}^2}}}} \right)} \,dt \hfill \\ \hfill \\ {\text{using}}\,\,\,partial\,\,fractions \hfill \\ \hfill \\ = \int_{}^{} {\,\left( {\frac{1}{4}\frac{1}{{1 + \sin t}} + \frac{1}{2}\frac{1}{{\,{{\left( {1 + \sin t} \right)}^2}}} - \frac{1}{4}\frac{1}{{\sin t - 1}}} \right)\,\cos tdt} \hfill \\ \hfill \\ Upon\,\,\,integrating \hfill \\ \hfill \\ \,Let,\,\,\,\,\,u = \sin t \hfill \\ \hfill \\ \int_{}^{} {\frac{{\sec t}}{{1 + \sin t}}\,dt} = \int_{}^{} {\left( {\frac{1}{4}\frac{1}{{u + 1}} + \frac{1}{2}\frac{1}{{\,{{\left( {u + 1} \right)}^2}}} - \frac{1}{4}\frac{1}{{u - 1}}} \right)} \cos tdt \hfill \\ \hfill \\ {\text{integrating}} \hfill \\ \hfill \\ = \frac{1}{4}\ln \left| {u + 1} \right| - \frac{1}{2}\frac{1}{{u + 1}} - \frac{1}{4}\ln \left| {u - 1} \right| - C \hfill \\ \hfill \\ = \frac{1}{4}\ln \left| {\frac{{u + 1}}{{u - 1}}} \right| - \frac{1}{2}\frac{1}{{u + 1}} + C \hfill \\ \hfill \\ Substitute\,\,back\,\,u = \sin t \hfill \\ \hfill \\ = \frac{1}{4}\ln \left| {\frac{{1 - \sin t}}{{1 + \sin t}}} \right| - \frac{1}{2}\frac{1}{{\sin t + 1}} + C \hfill \\ \end{gathered} \]
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