For `theta in (0,pi/2)`, prove that `int_0^theta"log"(1+tanthetatanx)dx=thetalog(sectheta)`

Explanation

`I=int_(0)^(theta)log(1+tan theta tan x) dx`
`=int_(0)^(theta)log(1+tan theta tan (theta-x))dx`
`=int_(0)^(theta)log(1+(tantheta(tan theta-tan x))/(1+tan theta tan x))dx`
`=int_(0)^(theta)"log"((1+tan^(2)theta))/(1+tan theta tan x)dx`
`=int_(0)^(theta)log(1+tan^(2)theta)dx-int_(0)^(theta)log(1+tan theta tan x)dx`
`=2theta log sec theta -I`
or `2I=2theta log sec theta`
or `I=thetalog sec theta`