cos x. sin2xx∈[0 cosx·Sin2xx∈[,丌] #1= cos x sin2 xsin2 xdx sin2 xd sin x- sin2 xd sin x 丌4 SIn- x SIn- x 03 3 3 2 2 2 2 0 sin sin sin sin xd x xd x    = −   5 5 2 2 2 2 2 2 4 sin sin 5 5 5 0 x x    = − = 2 2 3 3 2 2 0 cos sin cos sin I x xdx x xdx    =  −  故   3 2 3 2 cos sin [0, ) 2 cos sin [ , ] 2 x x x x x x         =   −   