Correct Answer - Option 4 : 4
\(\sqrt 2\)
Concept:
\(\rm {\sin ^2}{\rm{x}} + {\cos ^2}{\rm{x}} = 1\)
\(\rm\cos 2{\rm{x}} = cos^2{x} - sin^2{x}\)
Calculation:
Let, \({\rm{I}} = \mathop \smallint \limits_0^{2{\rm{\pi }}} \sqrt {1 + \cos \frac{{\rm{x}}}{2}} {\rm{dx}}\)
\(⇒ {\rm{I}} = \mathop \smallint \limits_0^{2{\rm{\pi }}} \sqrt {(\rm 2cos^2{(\frac {x}{4})}})\)
\(⇒ {\rm{I}} =\sqrt 2 \mathop \smallint \limits_0^{2{\rm{\pi }}} \cos{(\frac {\rm x}{4})}\)
\(⇒ {\rm{I}} = 4\sqrt 2\left[ { \sin {\rm{(\frac {x}{4}})}} \right]_0^{2\pi}\)
\(⇒ {\rm{I}} = 4\sqrt 2[ { \sin {\rm{(\frac {2\pi}{4}}) - sin0^o}}]\)
\(⇒ {\rm{I}} = 4\sqrt 2 (\sin {\frac {\pi}{2}})\)
\(⇒ {\rm{I}} = 4 \sqrt 2\)
Hence, option (4) is correct.