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What is \(\mathop \smallint \limits_0^{2{\rm{\pi }}} \sqrt {1 + \cos{ \frac{{\rm{x}}}{2}}} {\rm{dx\;}}\)equal to?
1. 8\(\sqrt 2\)
2. 2\(\sqrt 2\)
3. 6
4. 4\(\sqrt 2\)

1 Answer

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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.

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