∫(1/cos²x-1/sin²x)dx=∫((sin²x-cos²x)/(sin²x*cos²x))dx=
=-2*∫(cos(2x)/sin(2x))dx=-2*∫(ctg(2x))dx
Пусть u=2x  ⇒  du=2dx   dx=du/2
ctg(2x)=(1/2)*∫(ctg(u))du=(1/2)*∫(cos(u)/sin(u))du=
Пусть s=sin(u)   ds=cos(u)du   du=ds/cos(u)
=(1/2)*∫(1/s)ds=lns/2=ln(sin(u)/2=ln(sin(2x))/2.
-2*(ln(sin(2π/3))-ln(sin(π/3))/2=-(ln(√3/2)-ln(√3/2))=0.