三角换元
换元，x=tant=∫(π/4,π/3) d(tant)/(tan^2t√(1+tan^2t))=∫(π/4,π/3) (1/cos^2t)/(tan^2t*(1/cost)) dt=∫(π/4,π/3) cost/sin^2t dt=∫(π/4,π/3) sin^(-2)t d(sint)=-sin^(-1)t | (π/4,π/3)=2-2√3/3
