ज्ञात करे[ Evalute] ` int_(0)^(pi//4) log (1+tan x)dx`

Explanation

माना कि `f(x) = log (1+tanx)`
तो , ` f(pi/4-x) = log [ 1+ tan (pi/4 -x) = log (1+(1-tanx)/(1+tanx))`
या ` f (pi/4-x) = log. 2/(1+tanx) " "` …(2) ,
अब `(1) + (2) rArr f(x) (pi/4-x) = log (1+tan x) log (2/(1+tan x)) = log 2 `
` :. I = 1/2 int _(0)^(pi//4)[ f(x) + f (pi/4 -x)] dx = 1/2 int_(0)^(pi//4)log2 dx`
` = 1/2 * pi/4 log 2 = pi/8 log 2 `