Evaluate \(\rm\int_{0}^{\pi/2}\) log cot x dx .

Concept:

Definite Integral properties:

\(\mathop \smallint \limits_{\rm{0}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \mathop \smallint \limits_{\rm{0}}^{\rm{a}} {\rm{f}}\left( {{\rm{a}} - {\rm{x}}} \right){\rm{\;dx}}\)

Calculation :

Let , I = \(\rm\int_{0}^{π/2}\) log cot x dx             ....(i)

Now using property, \(\mathop \smallint \limits_{\rm{0}}^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{\;dx}} = \mathop \smallint \limits_{\rm{0}}^{\rm{a}} {\rm{f}}\left( {{\rm{a}} - {\rm{x}}} \right){\rm{\;dx}}\)

I = \(\rm\int_{0}^{π/2}\) log cot ( \(\rm \frac{\pi}{2}\) - x ) dx 

I = \(\rm\int_{0}^{π/2}\) log tan x dx                    .... (ii)

Adding eq. (i) and (ii), we get 

⇒ 2I = \(\rm\int_{0}^{π/2}\) (log cot x + log tan x ) dx 

⇒ 2I = \(\rm\int_{0}^{π/2}\) log ( tan x × cot x ) dx              [∵ log m + log n = log mn]

⇒ 2I = \(\rm\int_{0}^{π/2}\) log 1 dx 

⇒ 2I = 0                                                      [ ∵ log 1 = 0 ]   

⇒  I = 0 

The correct option is 1.