Question
The value of the integral $$\int\limits_0^{\frac{\pi }{2}} {\frac{{\sqrt {\cot \,x} }}{{\sqrt {\cot \,x} + \sqrt {\tan \,x} }}dx} $$ is-
A.
$$\frac{\pi }{4}$$
B.
$$\frac{\pi }{2}$$
C.
$$\pi $$
D.
none of these
Answer :
$$\frac{\pi }{4}$$
Solution :
$$\eqalign{
& I = \int_0^{\frac{\pi }{2}} {\frac{{\sqrt {\cot \,x} }}{{\sqrt {\cot \,x} + \sqrt {\tan \,x} }}dx} .....(1) \cr
& = \int_0^{\frac{\pi }{2}} {\frac{{\sqrt {\cot \,\left( {\frac{\pi }{2} - x} \right)} }}{{\sqrt {\cot \,\left( {\frac{\pi }{2} - x} \right)} + \sqrt {\tan \,\left( {\frac{\pi }{2} - x} \right)} }}dx} \cr
& I = \int_0^{\frac{\pi }{2}} {\frac{{\sqrt {tan\,x} }}{{\sqrt {\tan \,x} + \sqrt {cot\,x} }} = dx.....(2)} \cr
& {\text{Adding (1) and (2) we get}} \cr
& 2I = \int_0^{\frac{\pi }{2}} {\frac{{\sqrt {\cot \,x} + \sqrt {\tan \,x} }}{{\sqrt {\cot \,x} + \sqrt {\tan \,x} }} = dx = \int_0^{\frac{\pi }{2}} {1.dx} } \cr
& = \left( x \right)_0^{\frac{\pi }{2}} = \frac{\pi }{2} \cr
& \therefore I = \frac{\pi }{4} \cr} $$