Câu hỏi 12

\(I = \int\limits_{{\pi  \over 6}}^{{\pi  \over 4}} {{{dx} \over {{{\cos }^2}x{{\sin }^2}x}}}  = \int\limits_{{\pi  \over 6}}^{{\pi  \over 4}} {{{4dx} \over {{{\sin }^2}2x}}}  = \left. { - 2\cot 2x} \right|_{{\pi  \over 6}}^{{\pi  \over 4}} =  - 2\left( {0 - {1 \over {\sqrt 3 }}} \right) = {2 \over {\sqrt 3 }} = {{2\sqrt 3 } \over 3} \Rightarrow \left\{ \matrix{  a = 0 \hfill \cr   b = {2 \over 3} \hfill \cr}  \right. \Rightarrow a - b =  - {2 \over 3}\)

Chọn B.