Find the value of \[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\left( {\sqrt {\cot x} } \right)}}{{\left[ {\sqrt {\cot x} + \sqrt {\tan x} } \right]}}} dx\]
A. \[1\]
B. \[ - 1\]
C. \[\dfrac{\pi }{2}\]
D. \[\dfrac{\pi }{4}\]

Hint: The question is given in \[\cot x\] which we will be changing into \[\tan x\] to get two equations which will help us to solve the question in the easiest way.

Formula Used: The formula we will be using is \[\left[ {\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {\left( {a - x} \right)dx} } \right]\]

Complete step by step answer
Let us assume the equation given in the question as (i) equation .
\[I = \] \[\] \[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\left( {\sqrt {\cot x} } \right)}}{{\left[ {\sqrt {\cot x} + \sqrt {\tan x} } \right]}}} dx\] ----- (i)
Now we will be changing \[\cot x\] into \[\tan x\] by substituting \[x\] as \[\left( {\dfrac{\pi }{2} - x} \right)\] by applying the formula
\[I = \]\[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\cot \left( {\dfrac{\pi }{2} - x} \right)} }}{{\left[ {\sqrt {\cot \left( {\dfrac{\pi }{2} - x} \right)} + \sqrt {\tan \left( {\dfrac{\pi }{2} - x} \right)} } \right]}}} dx\]
\[\] \[\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\left( {\sqrt {\tan x} } \right)}}{{\left[ {\sqrt {\tan x} + \sqrt {\cot x} } \right]}}} dx\] ---- (ii)
 Now , Adding equation (i) and (ii), we will get
\[I + I = 2I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\cot x} + \sqrt {\tan x} }}{{\left[ {\sqrt {\tan x} + \sqrt {\cot x} } \right]}}} dx\] \[\]
\[2I = \int\limits_0^{\dfrac{\pi }{2}} {1dx} \]
Integrating R.H.S. of the above equation and we get
\[2I = {\left[ x \right]^{\dfrac{\pi }{2}}}_0\]
Substitute the upper and lower value, we get
\[2I = \dfrac{\pi }{2} - 0\]
Dividing both sides of the above equation by 2, we get
\[I = \dfrac{\pi }{4}\]
Hence the answer is D which is \[\dfrac{\pi }{4}\].

Note: This is the easiest way of solving this question because here we have used the basic formula of integration where we have changed \[\cot x\] to \[\tan x\] and vice versa to get two equations which directly helped us to get the values of the integration.