`int(1-cosx)/(cosx(1+cosx))dx`

Answer

`log|secx+tanx|-2tanx//2+C`

Explanation

`"Let " I=int(1-cosx)/(cosx(1+cosx))dx.`
`"Let " cosx=y. " Then "`
`(1-cosx)/(cosx(1+cosx))=(1-y)/(y(1+y))`
`=(1)/(y)-(2)/(1+y)`
`=(1)/(cosx)-(2)/(1+cosx)`
`:. I=int(1-cosx)/(cosx(1+cosx))dx`
`=int(1)/(cosx)dx-int(2)/(1+cosx)dx`
`=int sec x dx-int(2)/(2cos^(2)x//2)dx`
`=int secx-int sec^(2)x//2dx`
`=log|secx+tanx|-2tanx//2+C`