Biết (limits_0^{\pi }{3}} {\{{1 - \cos 2x}}{{1 + \cos 2x}}dx}  = a\sqrt 3  + \frac{\pi }{b}\)

\(\int\limits_0^{\frac{\pi }{3}} {\frac{{1 - \cos 2x}}{{1 + \cos 2x}}dx} = \int\limits_0^{\frac{\pi }{3}} {\frac{{2{{\sin }^2}x}}{{2{{\cos }^2}x}}} dx = \int\limits_0^{\frac{\pi }{3}} {\left( {\frac{1}{{{{\cos }^2}x}} - 1} \right)dx = \left. {\left( {\tan x - x} \right)} \right|_0^{\frac{\pi }{3}} = \sqrt 3 } - \frac{\pi }{3}\)

\( \Rightarrow \left\{ \begin{array}{l}a = 1\\b = - 3\end{array} \right. \Rightarrow a + b = - 2\).