secx的立方的原函数是什么?
发布网友
发布时间：2022-05-13 23:02
共1个回答
热心网友
时间：2023-11-04 20:00
=∫cosxdx/(cosx)^4
=∫d(sinx)/[1-(sinx)^2]^2, 令u=sinx
=∫/(1-u^2)^2
=0.25∫[1/(1-u)+1/(1+u)]^2
=0.25∫[1/(u-1)^2+1/(1+u)^2+2/(1-u)(1+u)]
=0.25∫[1/(u-1)^2+1/(u+1)^2+1/(1-u)+1/(1+u)]
=0.25[1/(1-u)-1/(1+u)-ln|1-u|+ln|1+u|]+C
=0.25[2u/(1-u^2)+ln|(1+u)/(1-u)|]+C
=0.5sinx/(cosx)^2+0.25ln|(1+sinx)/(1-sinx)|]+C
=0.5sinx(secx)^2+0.5ln|(1+sinx)secx|+C
