Answer
$$3\sqrt 3 - \pi $$
Work Step by Step
$$\eqalign{
& \int_0^\pi {{{\tan }^2}\frac{\theta }{3}} d\theta \cr
& {\text{Use the trigonometric identity}}\cr
& \tan^2 \theta + 1 = {\sec ^2}\theta \cr
& = \int_0^\pi {\left( {{{\sec }^2}\frac{\theta }{3} - 1} \right)} d\theta \cr
& {\text{We know that }}\cr
&\frac{d}{{dx}}\left[ {\tan x} \right] = {\sec ^2}x\cr
&{\text{ Then}}{\text{,}} \cr
& = \left( {3\tan \left( {\frac{\theta }{3}} \right) - \theta } \right)_0^\pi \cr
& {\text{Evaluate}} \cr
& = \left( {3\tan \left( {\frac{\pi }{3}} \right) - \pi } \right) - \left( {3\tan \left( {\frac{0}{3}} \right) - 0} \right) \cr
& {\text{Simplifying}} \cr
& = 3\left( {\sqrt 3 } \right) - \pi - 0 \cr
& = 3\sqrt 3 - \pi \cr} $$
