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Question

log (1 + tan x) dx 8.

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Solution

The function is given as,

y= 0 π 4 log( 1+tanx ) dx(1)

We have to calculate the integral of y.

From the property of integration,

0 b f( x ) dx= 0 b f( bx ) dx

y= 0 π 4 log( 1+tan( π 4 x ) ) dx

Using the formula of tan( bc )= tanbtanc 1+tanbtanc we can simplify the integral.

y= 0 π 4 log( 1+ tan π 4 tanx 1+tan π 4 tanx ) dx = 0 π 4 log( 1+ 1tanx 1+tanx ) dx = 0 π 4 log( 1tanx+1+tanx 1+tanx ) dx = 0 π 4 log( 2 1+tanx ) dx

Simplify further,

y= 0 π 4 [ log( 2 )log( 1+tanx ) ] dx = 0 π 4 [ log( 2 ) ] dx 0 π 4 log( 1+tanx ) dx (2)

We have to add equations (1) and (2) to get the solution,

2y= 0 π 4 log( 1+tanx ) dx+ 0 π 4 [ log( 2 ) ] dx 0 π 4 log( 1+tanx ) dx 2y= 0 π 4 [ log( 2 ) ] dx 2y=[ log( 2 ) ]× π 4 y= π 8 log2

Thus, the value of the integral is π 8 log2.


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