Question

求不定解积分：∫1/cos^3dx

积分没学好，劳烦大家了！步骤尽量详细啊，谢啦！

Answer 1

∫dx/cos³x
=∫sec³xdx
=∫secxdtanx
=secxtanx-∫tanxdsecx，分部积分法
=secxtanx-∫secxtan²xdx
=secxtanx-∫secx(sec²x-1)dx
=secxtanx-∫(sec³x-secx)dx
=secxtanx-∫sec³x+∫secxdx，两边都有重复表达式∫sec³xdx出现，移项到左边
2∫sec³xdx=secxtanx+∫secxdx，两边除以2
∫sec³xdx=(1/2)[secxtanx+ln|secx+tanx|]+C，secx的积分是ln|secx+tanx|

Answer 2

正解

Answer 3

∫1/cos^3xdx
=sinx/(2cos^2x)+1/2∫1/cosxdx
=sinx/(2cos^2x)+1/2[ln(secx+tgx)]+C

Answer 4

∫1/cos^3dx=∫cos^-3dx=-1/2sin^(-2)x