Question

Evaluate the following:

`int_0^(pi/2) log(tanx)dx`

Answer 1

Let I = `int_0^(pi/2) log(tanx)dx`

We use the property, `int_0^a f(x)dx = int_0^a f(a - x)dx`

Here, `a = pi/(2)`

Hence, changing x by `pi/(2) - x`, we get

I = `int_0^(pi/2) log[tan(pi/2 - x)]dx`

= `int_0^(pi/2) log(cotx)dx`

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

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

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

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

= – I

∴ 2I = 0

∴ I = 0