Question

Evaluate: `int_0^(pi/4) log(1 + tanx)  "d"x`

Answer 1

Let I = `int_0^(pi/4) log(1 + tanx)  "d"x`  ......(i)

= `int_0^(pi/4) log[1 + tan(pi/4 - x)]  "d"x`  ......`[∵ int_0^"a" "f"(x)  "d"x = int_0^"a" "f"("a" - x)  "d"x]`

= `int_0^(pi/4) log(1 + (tan  pi/4 - tan x)/(1 + tan  pi/4 * tanx)) "d"x `

= `int_0^(pi/4) log (1 + (1 - tan x)/(1 + tan x)) "d"x`

= `int_0^(pi/4) log ((1 + tanx + 1 - tan x)/(1 + tan x))  "d"x`

= `int_0^(pi/4) log(2/(1 + tan x)) "d"x`

= `int_0^(pi/4)[log2 - log(1 + tanx)]  "d"x`

= `log 2 int_0^(pi/4) 1* "d"x - int_0^(pi/4) log(1 + tanx)  "d"x`

∴ I = `log 2[x]_0^(pi/4) - "I"`   ......[From (i)]

∴ 2I = `log 2(pi/4 - 0)`

∴ I = `pi/8 log 2`