Please calculate the integral

limit is (0 to pi/4)

∫ln(1+tanx) dx

Dear student,
We have:
J = ∫ ln { 1 + tan [ (π/4) - x ] } dx ... on [0,π/4]

...= ∫ ln { 1 + [ ( 1 - tan x ) / ( 1 + tan x ) ] } dx

...= ∫ ln { 2 / ( 1 + tan x ) } dx

...= ∫ { ( ln 2 ) - ln ( 1 + tan x ) } dx ... on [0,π/4]

...= [ ( ln 2 ) ∫ dx ] - J .... from (2)
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∴ J = ( ln 2 ) { x on [0,π/4] } - J

∴ 2J = ( ln 2 ) [ (π/4) - 0 ]

∴ 2J = (π/4) ( ln 2 )

∴ J = ( π/8 )· ln 2 = (0.25)π· ln 2Regards
Sumit