Question

`int_0^(pi"/"4)` log(1 + tanθ) dθ = ______

Options

• `pi/4`log2

• `pi/4` log `1/2`

• `pi/8`log2

• `pi/8`log`1/2`

Answer 1

`int_0^(pi"/"4)` log(1 + tanθ) dθ = `underline(pi/8log2)`

Explanation:

Let I = `int_0^(pi"/"4)`log(1 + tanθ) dθ

= `int_0^(pi/4) log[1 + tan(pi/4 - theta)]`dθ .................`[∵ int_0^af(x)dx = int_0^af(a - x)dx]`

= `int_0^(pi/4) log(1 + (1 - tantheta)/(1 + tantheta))`dθ

= `int_0^(pi/4) log 2d theta - int_0^(pi/4) log(1 + tantheta)d theta`

∴ 2I = `int_0^(pi/4) log 2d theta ⇒ "I" = (log 2)/2 [theta]_0^(pi/4) = pi/8 log 2`