Question

Evaluate `int_0^(π//4) log (1 + tanx)dx`.

Answer 1

Let I = `int_0^(π//4) log (1 + tanx)dx`  ...(i)

By using property

`int_0^a f(x) = int_0^a f(a - x)`

I = `int_0^(π//4) log [1 + tan(π/4 - x)]dx`

= `int_0^(π//4) log [1 + (tan  π/4 - tan x)/(1 + tan  π/4 tan x)]dx`

= `int_0^(π//4) log [1 + (1 - tanx)/(1 + tanx)]dx`

= `int_0^(π//4) log [2/(1 + tanx)]dx`

= `int_0^(π//4) log 2 - int_0^(π//4) log (1 + tanx)dx`  ...(ii)

On adding equations (i) and (ii),

2I = `int_0^(π//4) log (1 + tanx)dx + int_0^(π//4) log 2 dx - int_0^(π//4) log (1 + tanx)dx`

`\implies` 2I = `int_0^(π//4) log 2 dx` 

`\implies` 2I = `log 2 int_0^(π//4) 1.dx`

`\implies` 2I = `log 2 [x]_0^(π//4)`

`\implies` I = `log2/2 xx π/4 = π/8 log 2`