Correct Answer - Option 1 :
\(\frac{\pi}{4}\)
Concept:
Property of definite integrals:
\(\mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {{\rm{a}} - {\rm{x}}} \right){\rm{dx}}\)
Calculation:
Given:
\(I=\int^{\frac {\pi} 2}_0 \frac {1}{1 + \sqrt {\cot x}}dx\)
\(I=\int^{\frac {\pi} 2}_0 \frac {1}{1 + \sqrt {\frac{cos\ x}{sin\ x}}}dx\)
\(I=\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}{\rm{dx}}\) --- (1)
Using the property of definite integrals:
\(\mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \mathop \smallint \nolimits_0^{\rm{a}} {\rm{f}}\left( {{\rm{a}} - {\rm{x}}} \right){\rm{dx}}\)
\({\rm{I}} = \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{\sqrt {\sin \left( {\frac{{\rm{\pi }}}{2} - {\rm{x}}} \right)} }}{{\sqrt {\sin \left( {\frac{{\rm{\pi }}}{2} - {\rm{x}}} \right)} + \sqrt {\cos \left( {\frac{{\rm{\pi }}}{2} - {\rm{x}}} \right)} }}{\rm{dx}} \)
\(I= \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{\sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}{\rm{dx\;}}\) --- (2)
Adding equation (1) and (2), we get
\(2I=\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}{\rm{dx}}+ \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{\sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}{\rm{dx\;}}\)
\(2I = {\rm{\;}}\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} \frac{{\sqrt {\sin x} + \sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}{\rm{dx}}\)
\( 2I= \mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{2}} 1{\rm{dx}}\)
\(2I = {\rm{\;}}\left[ x \right]_0^{\frac{{\rm{\pi }}}{2}} \)
\(2I= \frac{{\rm{\pi }}}{2}\)
\(\therefore I= \frac{{\rm{\pi }}}{4}\)
