例12求∫ SIn-cos 解 sIn—cos sInx 2 =csc xdx =-4cotx +C例12 . 2 cos 2 sin 1 2 2 x x x d 求  解 x x x d  2 cos 2 sin 1 2 2 x x d        = 2 2 sin 1 xdx  = 2 4csc = −4cot x +C