Question

log (1 + tan x) dx 8.

Answer 1

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( b−x ) dx

y =  ∫ 0 π 4 log( 1+tan( π 4 −x ) ) dx

Using the formula of tan( b−c )= tanb−tanc 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+ 1−tanx 1+tanx ) dx = ∫ 0 π 4 log( 1−tanx+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.