\(\mathop \smallint \limits_0^{2{\rm{\pi }}} \sqrt {1 + \cos{ \frac{{\rm{x}}}{2}}} {\rm{dx\;}}\) किसके बराबर है?

संकल्पना:

\(\rm {\sin ^2}{\rm{x}} + {\cos ^2}{\rm{x}} = 1\)

\(\rm\cos 2{\rm{x}} = cos^2{x} - sin^2{x}\)

गणना:

माना कि \({\rm{I}} = \mathop \smallint \limits_0^{2{\rm{\pi }}} \sqrt {1 + \cos \frac{{\rm{x}}}{2}} {\rm{dx}}\)  है। 

\(⇒ {\rm{I}} = \mathop \smallint \limits_0^{2{\rm{\pi }}} \sqrt {(\rm 2cos^2{(\frac {x}{4})}})\)

\(⇒ {\rm{I}} = \sqrt 2\mathop \smallint \limits_0^{2{\rm{\pi }}} \cos{(\frac {\rm x}{4})}\)

\(⇒ {\rm{I}} = 4\sqrt 2\left[ { \sin {\rm{(\frac {x}{4}})}} \right]_0^{2\pi}\)

\(⇒ {\rm{I}} = 4\sqrt 2[ { \sin {\rm{(\frac {2\pi}{4}}) - sin0^o}}]\)

\(⇒ {\rm{I}} = 4 \sqrt 2(\sin {\frac {\pi}{2}})\)

\(⇒ {\rm{I}} = 4 \sqrt 2\)

अतः विकल्प (4) सही है।