$\int\limits_0^\pi {\left[ {\cot x} \right]} dx$where [.] denotes the greatest integer function, is equal to
$
  {\text{A}}{\text{. }}\dfrac{\pi }{2} \\
  {\text{B}}{\text{. 1}} \\
  {\text{C}}{\text{. - 1}} \\
  {\text{D}}{\text{. - }}\dfrac{\pi }{2} \\
 $

Hint: To solve this question we have to understand property of greater integer function like: $
  \left[ x \right] + \left[ { - x} \right] = 1,x \notin I \\
  \left[ x \right] + \left[ { - x} \right] = 0,x \in I \\
$
Now we have to consider cot function on which x it gives integer value and on which other value and by doing this we can proceed further in the question.

Complete step-by-step answer:
In order to use the property,
$
  \left[ x \right] + \left[ { - x} \right] = - 1,x \notin I \\
  \left[ x \right] + \left[ { - x} \right] = 0,x \in I \\
$
We need to apply the below property of integers first-
$\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^b {f\left( {b - a - x} \right)dx} $
This gives us:
$
  \int\limits_0^\pi {\left[ {\cot x} \right]dx} = \int\limits_0^\pi {\left[ {\cot \left( {\pi - x} \right)} \right]dx} \\
  \int\limits_0^\pi {\left[ {\cot \left( {\pi - x} \right)} \right]dx} = \int\limits_0^\pi {\left[ { - \cot x} \right]dx} \\
$
This shows:
$
  I = \int\limits_0^\pi {\left[ {\cot x} \right]dx} \\
  I = \int\limits_0^\pi {\left[ {\cot \left( {\pi - x} \right)} \right]dx} = \int\limits_0^\pi {\left[ { - \cot x} \right]dx} \\
$
Adding the two integrals-
$
  2I = \int\limits_0^\pi {\left[ {\cot x} \right]dx} + \int\limits_0^\pi {\left[ { - \cot x} \right]dx} \\
  2I = \int\limits_0^\pi {\left( {\left[ {\cot x} \right] + \left[ { - \cot x} \right]} \right)dx} \\
$
Now in the interval $0 - \pi $ cot(x) is not an integer so, from the property of greatest integer function we get,
 $
  2I = \int\limits_0^\pi {\left( {\left[ {\cot x} \right] + \left[ { - \cot x} \right]} \right)dx} \\
  2I = \int_0^\pi { - 1.dx} \\
  2I = - \left( {\pi - 0} \right) \\
  2I = - \pi \\
  I = \dfrac{{ - \pi }}{2} \\
$
The correct option is- D.

Note: Whenever we get this type of question the key concept of solving is we have to care more about negative values because on taking greater integers some students confuse. And we have to remember properties of definite integration like $\int\limits_a^b {f\left( x \right)dx} = \int\limits_a^b {f\left( {b - a - x} \right)dx} $.