求不定积分∫[1/sin^2cos^2 (x)]dx
问题描述：
求不定积分∫[1/sin^2cos^2 (x)]dx
答
∫[1/sin²xcos²x]dx
=∫[4/(2sinxcosx)²]dx
=∫[4/sin²2x]dx
=∫2csc²2xd2x
=-2∫(-csc²2x)d2x
=-2cot2x+C