高数,这道题怎么做呀,麻烦写下过程谢谢

2025-01-06 04:55:20
推荐回答(1个)
回答1:

令x=asect,dx=ad(sect)
原式=∫atant*ad(sect)
=a^2*∫tantd(sect)
=a^2*[tantsect-∫sec^3tdt]
∫sec^3tdt=∫sectd(tant)
=secttant-∫tan^2tsectdt
=secttant-∫(sec^2t-1)sectdt
=secttant-∫sec^3tdt+∫sectdt
=secttant+ln|sect+tant|-∫sec^3tdt
所以∫sec^3tdt=(1/2)*[secttant+ln|sect+tant|]+C
原式=a^2*[tantsect-∫sec^3tdt]
=(a^2/2)*[tantsect-ln|sect+tant|]+C
=(a^2/2)*[√(x^2-a^2)/x-ln|x/a+√(x^2-a^2)/a|]+C
=(a^2/2)*[√(x^2-a^2)/x-ln|x+√(x^2-a^2)|]+C,其中C是任意常数