令tanx=t,则x=arctant,dx=dt/(1+t^2),cos^2x=1/(1+t^2),sin^2x=1-cos^2t=t^2/(1+t^2)∫dx/=∫dx/[1+sin²(2x)]=∫dx/(1+4sin²xcos²x)=∫[dt/(1+t^2)]/[(1+4t^2/(1+t^2)^2]=∫(1+t^2)/(t^4+6t^2+1)dt=...