Question

 Show that: `int _0^(pi/4) log (1 + tanx) dx = pi/8 log2`

Answer 1

Let I = `int _0^(pi/4) log (1 + tan x) dx`

= `int _0^(pi/4) log {1+ tan(pi/4 - x)}dx`   ......`(∵  int _0^a f(x) dx = int  f (a  - x) dx) `

= `int _0^(pi/4) log {1 + ((tan pi/4 - tanx))/(1 + tan pi/4 tanx}}dx`

= `int _0^(pi/4) {1 + (1 - tanx)/(1 + tanx)} dx`

 = `int _0^(pi/4) log{(1 + tanx + 1 - tanx)/(1 + tanx)} dx`

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

= `int_0^(pi/4) {log2 - log(1 + tanx)} dx`

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

I = `log2[x]_0^(pi/4) - I`

2I = `log2[pi/4 - 0]`

I = `pi/8 . log2`

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