cos 2x /sin^2 x*cos^2 x不定积分

∫cos2xdx/(sin^2xcos^2x)
= 4∫cos2xdx/(2sinxcosx)^2
=4 ∫cos2xdx/(sin2x)^2
=2 ∫cos2xd(2x)/(sin2x)^2
=2 ∫d(sin2x)/(sin2x)^2
=-2*1/(sin2x)+c
=-2csc2x+c.答案是“-tanx-cotx+c"我要过程....拜托您了因为-tanx-cotx=-(sinx/cosx+cosx/sinx)=-(sin^2x+cos^2x)/sinxcosx=-1/sinxcosx=-2/sin2x=-2csc2x.或者采取如下积分：∫cos 2x dx/sin^2 x*cos^2 x=∫(2cos^2x-1)dx/sin^2x*cos^2x=∫（cos^2 x-sin^2 x)dx/(sin^2 x*cos^2 x)= ∫（1/sin^2 X-1/cos^2 x)dx=∫dx/sin^2x-∫dx/cos^2x= ∫csc^2xdx-∫sec^2xdx=-cotx-tanx+c.