The correct option is A ln (2).π8
I want to say that this is a very famous question. It’s due to the fact that it emphasises the importance of the following property for its evaluation.
∫a0f(x)dx=∫a0f(a−x).dx.
So understand all the properties of Definite Integrals and you can use them as an essential tool-kit to tackle this chapter.
So here, let the integral to be evaluated be `I'.
∴∫π40ln(1+tanx)dx.
=∫π40ln(1+tan(π4−x))dx.
=∫π40ln(1+1−tan(x)1+tan(x))dx
=∫π40ln(21+tan(x))dx
=∫π40ln(2)dx−∫π/40ln(1+tan(x))dx
=ln(2)x|π40−l
2l=In (2).π4
I=In (2).π8