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: 89

Answer

\[ = - \frac{2}{{1 + \tan \,\left( {\frac{x}{2}} \right)}} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{dx}}{{1 + \sin x}}} \hfill \\ \hfill \\ se\,t = 2{\tan ^{ - 1}}\,\left( u \right)\,\,\,\,\,then\,\,dx = 2\left( {\frac{{du}}{{1 + {u^2}}}} \right) \hfill \\ and\,\,\sin x = \frac{{2u}}{{1 + {u^2}}} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \int_{}^{} {\frac{{dx}}{{1 + \sin x}}} = \int_{}^{} {\frac{{\frac{{2du}}{{1 + {u^2}}}}}{{1 + \frac{{2u}}{{1 + {u^2}}}}}} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \int_{}^{} {\frac{2}{{\,{{\left( {1 + u} \right)}^2}}}du} \hfill \\ \hfill \\ {\text{integrate}} \hfill \\ \hfill \\ = - \frac{2}{{1 + u}} + C \hfill \\ \hfill \\ substitute\,\,for\,\,\,u \hfill \\ \hfill \\ = - \frac{2}{{1 + \tan \,\left( {\frac{x}{2}} \right)}} + C \hfill \\ \end{gathered} \]
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