Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Practice Exercises - Page 439: 39

Answer

$$\ln 8$$

Work Step by Step

$$\eqalign{ & \int_0^\pi {\tan \frac{x}{3}} dx \cr & {\text{Use the identity tan}}\theta = \frac{{\sin \theta }}{{\cos \theta }} \cr & \int_0^\pi {\frac{{\sin \left( {x/3} \right)}}{{\cos \left( {x/3} \right)}}} dx \cr & {\text{Use substitution:}}\cr &{\text{Let }}u = \cos \left( {x/3} \right),\cr &{\text{ so that }}du = - \frac{1}{3}\sin \left( {x/3} \right)dx \cr & {\text{The new limits on }}u{\text{ are found as follows}} \cr & \,\,\,\,\,\,{\text{If }}x = \pi,{\text{ then }}u = \cos \left( {\pi /3} \right) = 1/2 \cr & \,\,\,\,\,\,{\text{If }}x = 0,{\text{ then }}u = \cos \left( 0 \right) = 1 \cr & {\text{Then}} \cr & \int_0^\pi {\frac{{\sin \left( {x/3} \right)}}{{\cos \left( {x/3} \right)}}} dx = - 3\int_1^{1/2} {\frac{1}{u}} du \cr & {\text{integrate}} \cr & = - 3\left( {\ln \left| u \right|} \right)_1^{1/2} \cr & {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr & = - 3\left( {\ln \left| {1/2} \right| - \ln 1} \right) \cr & {\text{simplifying, we get:}} \cr & = - 3\left( { - \ln 2} \right) \cr & = 3\ln 2 \cr & = \ln 8 \cr} $$
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