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( 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.

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