Question

`int_0^{pi/2} log(tanx)dx` = ______

Options

• 0

• 1

• `pi/4`

• `pi/2`

Answer 1

`int_0^{pi/2} log(tanx)dx` = 0

Explanation:

Let I = `int_0^{pi/2} log(tanx)dx` ...................(i)

∴ I = `int_0^{pi/2} log[tan(pi/2 - x)]dx` ..............`[∵ int_0^a f(x)dx = int_0^a f(a - x)dx]`

∴ I = `int_0^{pi/2} log(cotx)dx` ...................(ii)

Adding (i) and (ii), we get

2I = `int_0^{pi/2}[log(tanx) + log(cotx)]dx`

= `int_0^{pi/2}log(tanx.cotx)dx`

= `int_0^{pi/2} (log1)dx`

= `int_0^{pi/2} 0 dx`

∴ I = 0