Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.6 Other Integration Strategies - 7.6 Exercises: 33

Answer

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

Work Step by Step

\[\begin{gathered} \int_{}^{} {\frac{{\cos x}}{{{{\sin }^2}x + 2\sin x}}\,dx} \hfill \\ \hfill \\ set\,\,u = \sin x\,\,\,\,then\,\,\,\,du = \cos xdx \hfill \\ \hfill \\ \int_{}^{} {\frac{{\cos x}}{{{{\sin }^2}x + 2\sin x}}\,dx} = \int_{}^{} {\frac{{du}}{{{u^2} + 2u}}} = \int_{}^{} {\frac{{du}}{{u\,\left( {u + 2} \right)}}} \hfill \\ \hfill \\ use\,\,the\,\,formula\,\,\int_{}^{} {\frac{1}{{u\,\left( {a + b} \right)}}du = \frac{1}{a}\ln \left| {\frac{u}{{a + bu}}} \right| + C} \hfill \\ then \hfill \\ \hfill \\ = \frac{1}{2}\ln \left| {\frac{u}{{u + 2}}} \right| + C \hfill \\ \hfill \\ substitute\,\,back\,\,u = \sin x \hfill \\ \hfill \\ = \frac{1}{2}\ln \left| {\frac{{\sin x}}{{\sin x + 2}}} \right| + C \hfill \\ \hfill \\ \end{gathered} \]
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