第一型曲线积分,这道题怎么求

2025-01-21 06:27:58
推荐回答(1个)
回答1:

设x=tant =>dx=d(tant)=sec²tdt ∴ ∫(1/√(1+x^2))dx =∫(1/sect)sec²tdt =∫sectdt =∫cost/(cost)^2 dt =∫1/(cost)^2 dsint =∫1/(1-(sint)^2) dsint 令sint = θ化为∫1/(1-θ^2)dθ=(ln|1+x|-ln|1-x|)/2+C =ln(√((1+θ)/(1-θ)))+C =ln|sect+tant|+C =lnl√(1+tan^2t)+tantl+c =lnl√(1+x^2)+xl+c