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