第1个回答 2020-05-12
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
=∫(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
相似回答
