## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 13 - The Trigonometric Functions - 13.3 Integrals of Trigonometric Functions - 13.3 Exercises - Page 697: 13

#### Answer

$$- \ln \left| {1 + \cos x} \right| + C$$

#### Work Step by Step

\eqalign{ & \int {\frac{{\sin x}}{{1 + \cos x}}} dx \cr & {\text{set }}u = 1 + \cos x{\text{ then }}\frac{{du}}{{dx}} = - \sin x,\,\,\,\,\,\,\,\,\frac{{ - du}}{{\sin x}} = dx \cr & {\text{write the integrand in terms of }}u \cr & \int {\frac{{\sin x}}{{1 + \cos x}}} dx = \int {\frac{{\sin x}}{u}} \left( {\frac{{ - du}}{{\sin x}}} \right) \cr & {\text{cancel common factor}} \cr & = \int {\frac{1}{u}} \left( { - du} \right) \cr & = - \int {\frac{1}{u}} du \cr & {\text{integrate by using the integration rule }}\int {\frac{1}{x}} dx = \ln \left| x \right| + C \cr & {\text{then}} \cr & = - \ln \left| u \right| + C \cr & {\text{write in terms of }}x \cr & = - \ln \left| {1 + \cos x} \right| + C \cr}

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