∫（csc2x-cot2x)^2

Question

如题所述

Answer 1

解：原式=∫[1/sin(2x)-cos(2x)/sin(2x)]²dx
=∫[(1-cos(2x))/sin(2x)]²dx
=∫[(2sin²x)/(2sinxcosx)]²dx
=∫(sinx/cosx)²dx
=∫tan²xdx
=∫(sec²x-1)dx
=∫sec²xdx-∫dx
=∫d(tanx)-∫dx
=tanx-x+C (C是积分常数)。