A) 1
B) \[-1\]
C) \[\frac{\pi }{2}\]
D) \[\frac{\pi }{4}\]
Correct Answer: D
Solution :
Key Idea: \[\int_{0}^{a}{f(x)}dx=\int_{0}^{a}{f(a-x)}dx\] Let \[I=\int_{0}^{\pi /2}{\frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}}}dx\] ...(i) \[\Rightarrow \]\[I=\int_{0}^{\pi /2}{\frac{\sqrt{\cot \left( \frac{\pi }{2}-x \right)}}{\sqrt{\cot \left( \frac{\pi }{2}-x \right)}+\sqrt{\tan \left( \frac{\pi }{2}-x \right)}}}dx\] \[\Rightarrow \]\[I=\int_{0}^{\pi /2}{\frac{\sqrt{\tan x}}{\sqrt{\tan x}+\sqrt{\cot x}}}dx\] ?. (ii) On adding Eqs. (i) and (ii), we get \[2I=\int_{0}^{\pi /2}{\frac{\sqrt{\tan x}+\sqrt{\cot x}}{\sqrt{\tan x}+\sqrt{\cot x}}}dx\] \[=\int_{0}^{\pi /2}{dx}\frac{\pi }{2}-0\] \[\Rightarrow \] \[2I=\frac{\pi }{2}\Rightarrow I=\frac{\pi }{4}\] Alternative Method Let \[I=\int_{0}^{\pi /2}{\frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}}}dx\] \[\Rightarrow \]\[I=\int_{0}^{\pi /2}{\frac{\sqrt{\cos x/\sin x}}{\sqrt{\cos x/\sin x}+\sqrt{\sin x/\cos x}}}dx\] \[\Rightarrow \]\[I=\int_{0}^{\pi /2}{\frac{\sqrt{\cos x}}{\sqrt{\sin x}}\times \frac{\sqrt{\sin x}.\sqrt{\cos x}}{(\cos x+\sin x)}}dx\] \[\Rightarrow \]\[I=\int_{0}^{\pi /2}{\frac{\cos x}{\cos x+\sin x}}dx\] ?.(i) \[\Rightarrow \]\[I=\int_{0}^{\pi /2}{\frac{\cos \left( \frac{\pi }{2}-x \right)}{\cos \left( \frac{\pi }{2}-x \right)+\sin \left( \frac{\pi }{2}-x \right)}}dx\] \[=\int_{0}^{\pi /2}{\frac{\sin x}{\sin x+\cos x}}\] ?.(ii) On adding Eqs. (i) and (ii), we get \[2I=\int_{0}^{\pi /2}{dx}\] \[\Rightarrow \] \[2I=\frac{\pi }{2}\] \[\Rightarrow \] \[I=\frac{\pi }{4}\]
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