∫1/(sec^2x+1)dx等于多少?

来源：学生作业帮编辑：作业帮分类：综合作业时间：2024/05/12 14:50:51

∫ 1/(sec²x + 1) dx
= ∫ [(1 + sec²x) - sec²x]/(sec²x + 1) dx
= ∫ dx - ∫ sec²x/(sec²x + 1) dx
= x - ∫ d(tanx)/(tan²x + 2),恒等式1 + tan²x = sec²x
= x - (1/√2)arctan[(tanx)/√2] + C

∫1/(sec^2x+1)dx等于多少? ∫secx/sec^2x-1 dx ∫1/根号x*sec^2(1-根号x)dx ∫sec²x/1+tanx dx 1、∫[2/(1+x^2) -sec^2)dx ∫[(sec^2x-1)secx]dx= ∫ sec^2 x dx =∫dx/(2cos^2(x/2)+1)=∫sec^(x/2)dx/(2+sec^2(x/2))怎么出来的? ∫（sec^2x+sinx)dx ∫(1+cos^3 x)sec^2 x dx 求不定积分~ ∫sec^4x dx ∫sec^2x tan^2x dx ∫tan^2xdx=∫(sec^2x-1)dx