Evaluate: π40log(1+tan(x))dx ?

The answer is π2, but how?

1 Answer
Nov 14, 2017

π40log(1+tanx)dx=π8log2

Explanation:

As a0f(x)dx=a0f(ax)dx

Let I=π40log(1+tanx)dx=π40log(1+tan(π4x))dx

= π40log(1+1tanx1+tanx)dx

= π40log(21+tanx)dx

= π40log2dxπ40log(1+tanx)dx

= π40log2dxI

Hence 2I=π40log2dx

and I=log2×π4×12=π8log2