Evaluate `int_0^(pi/4) log(1+tanx)dx`

Explanation

` L.H.S. = I = int_(0)^(pi//4) log (1 + tanx)dx`
` = int_(0)^(pi//4) log{1 + tan ((pi)/(4)- x)} dx `
` [ because int_(0)^(a) f (t) dx = int_(0)^(a) f (a - x) dx] `
`= int_(0)^(pi//4) log{1 + ((tan""(pi)/(4)- tan x))/(1+tan""(pi)/(4) tan x)}dx`
` = int_(0)^(pi//4) log [ 1 + (1- tanx)/(1+ tanx)] dx `
` = int_(0)^(pi//4) log [ (1 + tan x + 1 - tan x)/(1 + tan x)] dx `
`= int_(0)^(pi//4)log ((2)/(1 + tanx)) dx `
`=int_(0)^(pi//4) [ log2 - log (1 + tanx)]dx `
`= int_(0)^(pi//4) log2*dx - int_(0)^(pi//4) log (1 + tanx)* dx `
` therefore I = log 2[x]_(0)^(pi//4) - 1`
` therefore 2I = log2 [(pi)/(4)] `
` therefore I = (pi)/(8) log 2 = R.H.S " "` Hence Proved