# Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.2 Derivatives And Integrals Involving Logarithmic Functions - Exercises Set 6.2 - Page 426: 72

$$2\ln 2 - \ln 3$$

#### Work Step by Step

\eqalign{ & \int_0^{\pi /3} {\frac{{\sin x}}{{1 + \cos x}}dx} \cr & {\text{Set }}u = 1 + \cos x \to du = - \sin xdx \cr & u = 1 + \cos x,{\text{ for }}x = \pi /3,\,\,u = 3/2 \cr & u = 1 + \cos x,{\text{ for }}x = 0,\,\,u = 2 \cr & {\text{using the substitution}} \cr & \int_0^{\pi /3} {\frac{{\sin x}}{{1 + \cos x}}dx} = - \int_2^{3/2} {\frac{1}{u}} du \cr & {\text{integrate}} \cr & = - \left[ {\ln \left| u \right|} \right]_2^{3/2} \cr & {\text{evaluate}} \cr & = - \left[ {\ln \left| {\frac{3}{2}} \right| - \ln \left| 2 \right|} \right] \cr & = 2\ln 2 - \ln 3 \cr}

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