`int_(0)^((pi)/(4))log(1+tan x)dx`

Explanation

माना `I=int_(0)^(pi//4)log(1+tanx)dx" …(1)"`
`=int_(0)^(pi//4)log{1+tan((pi)/(4)-x)}dx`
[ प्रगुण (4 ) से ]
`=int_(0)^(pi/4)log{1+(tan.(pi)/(4)-tanx)/(1+tan.(pi)/(4)tanx)}dx`
`=int_(0)^(pi//4)log{1+(1-tanx)/(1+tanx)}dx`
`=int_(0)^(pi//4)log((2)/(1+tanx))dx`
`=int_(0)^(pi//4)log2. dx - int_(90)^(pi//4)log(1+tanx)dx`
`rArr" "I=log 2 [x]_(0)^(pi//4)-I`
`rArr" "2I=(pi)/(4)log2 " "rArr" "I=(pi)/(8)log2`