Question

Evaluate: `int_0^(pi/4)log(1+tan(x))*dx` ?

The answer is `pi/2`, but how?

Answer 1

`int_0^(pi/4)log(1+tanx)dx=pi/8log2`

Explanation:

As `int_0^af(x)dx=int_0^af(a-x)dx`

Let `I=int_0^(pi/4)log(1+tanx)dx=int_0^(pi/4)log(1+tan(pi/4-x))dx`

= `int_0^(pi/4)log(1+(1-tanx)/(1+tanx))dx`

= `int_0^(pi/4)log(2/(1+tanx))dx`

= `int_0^(pi/4)log2dx-int_0^(pi/4)log(1+tanx)dx`

= `int_0^(pi/4)log2dx-I`

Hence `2I=int_0^(pi/4)log2dx`

and `I=log2xxpi/4xx1/2=pi/8log2`