Câu hỏi 52

\(\eqalign{  & a = \int\limits_0^{{\pi  \over 2}} {{{4{{\sin }^3}x} \over {1 + \cos x}}dx}  = 4\int\limits_0^{{\pi  \over 2}} {{{\left( {1 - {{\cos }^2}x} \right)\sin x} \over {1 + \cos x}}dx}  = 4\int\limits_0^{{\pi  \over 2}} {\left( {1 - \cos x} \right)\sin xdx}  =  - 4\int\limits_0^{{\pi  \over 2}} {\left( {1 - \cos x} \right)d\left( {\cos x} \right)}   \cr   & \,\,\, =  - \left. {4\left( {\cos x - {{{{\cos }^2}x} \over 2}} \right)} \right|_0^{{\pi  \over 2}} = 2  \cr   & b = \int\limits_{{\pi  \over 2}}^{{\pi  \over 3}} {\left( {\sin 2x + \cos x} \right)dx}  = \left. {\left( { - {{\cos 2x} \over 2} + \sin x} \right)} \right|_{{\pi  \over 2}}^{{\pi  \over 3}} = {{1 + 2\sqrt 3 } \over 4} - {3 \over 2} = {{2\sqrt 3  - 5} \over 4}  \cr   &  \Rightarrow P = a + 2b\sqrt 3  = {{10 - 5\sqrt 3 } \over 2} \Rightarrow \left\{ \matrix{  m = 10 \hfill \cr   n = 5 \hfill \cr}  \right. \Rightarrow m - n = 5 \cr} \)

Chọn B.