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Please calculate the integral limit is (0 to pi/4) ∫ln(1+tanx) dx

Please calculate the integral
limit is (0 to pi/4)
∫ln(1+tanx) dx

Grade:11

1 Answers

Sumit Majumdar IIT Delhi
askIITians Faculty 137 Points
8 years ago
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 2I=\int_{0}^{\frac{\pi }{4}} ln\left ( 1+tanx \right )dx=\int_{0}^{\frac{\pi }{4}} ln\left ( 1+tan\left (\frac{\pi }{4}-x \right ) \right )dx=\int_{0}^{\frac{\pi }{4}} ln\left ( 1+ \frac{1-tan\left ( x \right )}{1+tan\left ( x \right )}\right )dx=\int_{0}^{\frac{\pi }{4}} ln\left ( \frac{2}{1+tan\left ( x \right )}\right )dx=\int_{0}^{\frac{\pi }{4}} \left (ln\left ( 2 \right ) \right -ln\left ( 1+tan\left ( x \right ) \right ))dx=\int_{0}^{\frac{\pi }{4}} \left ( ln2 \right )dx-I\Rightarrow I=\frac{1}{2}\int_{0}^{\frac{\pi }{4}} \left ( ln2 \right )dx=\frac{\pi }{8}ln2Regards
Sumit

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