Answer
$$3\ln 2$$
Work Step by Step
$$\eqalign{
& \int_0^\pi {\tan \frac{\theta }{3}} d\theta \cr
& {\text{Rewrite}} \cr
& \int_0^\pi {\tan \frac{\theta }{3}} d\theta = \int_0^\pi {\frac{{\sin \left( {\theta /3} \right)}}{{\cos \left( {\theta /3} \right)}}} d\theta \cr
& {\text{Let }}u = \cos \left( {\theta /3} \right),{\text{ }}du = - \sin \left( {\frac{\theta }{3}} \right)\left( {\frac{1}{3}} \right)d\theta ,{\text{ }}\sin \left( {\frac{\theta }{3}} \right)d\theta = - 3du \cr
& {\text{The new limits of integration are}} \cr
& x = \pi \to u = \cos \left( {\frac{\pi }{3}} \right) = \frac{1}{2} \cr
& x = 0 \to u = \cos \left( {\frac{0}{3}} \right) = 1 \cr
& {\text{Substitute and integrate}} \cr
& \int_0^\pi {\frac{{\sin \left( {\theta /3} \right)}}{{\cos \left( {\theta /3} \right)}}} d\theta = \int_1^{1/2} {\frac{{ - 3du}}{u}} \cr
& = - 3\int_1^{1/2} {\frac{{du}}{u}} \cr
& {\text{Integrating}} \cr
& = - 3\left[ {\ln \left| u \right|} \right]_1^{1/2} \cr
& {\text{Evaluating}} \cr
& {\text{ = }} - 3\left( {\ln \frac{1}{2} - \ln 1} \right) \cr
& = - 3\ln \left( {\frac{1}{2}} \right) \cr
& = 3\ln 2 \cr} $$
