设t=tanx,则x=arctant,dx=dt/(1+t²),sec²x=1+t²
故∫sin²x/(1+cos²x)dx=∫tan²x/(1+sec²x)dx
=∫t²/[(1+t²)(2+t²)]dt
=∫[2/(2+t²)-1/(1+t²)]dt
=√2∫d(t/√2)/[1+(t/√2)²]-∫dt/(1+t²)
=√2arctan(t/√2)-arctant+C (C是积分常数)
=√2arctan(tanx/√2)-x+C.