Question

求不定积分(1/sin^2xcos^2x)dx

Answer 1

原式=∫4dx/(2sinxcosx)²
=4∫dx/sin²2x
=2∫csc²2xd2x
=-2cot2x+C

Answer 2

原式=∫4/4sin^2xcos^2xdx
=∫4/sin^2(2x)dx
=2∫csc^2(2x)d(2x)
=-2ctg(2x)+C