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

Concept:

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

Calculation:

 Given: \({\rm{I}} = \mathop \smallint \limits_0^{2{\rm{\pi }}} √ {1 + \cos 2x} {\rm{dx}}\)

\(= \mathop \smallint \limits_0^{2{\rm{\pi }}} √ {{{\sin }^2}\ x + {{\cos }^2}\ x + \cos^2 x - \sin^2x}\)                  (1 = \({\sin ^2}x + {\cos ^2}x\))

\(= \mathop \smallint \limits_0^{2{\rm{\pi }}} √ {2 \cos^2 x}\ dx\)             

= \(\rm √ 2 \mathop \smallint \limits_0^{2{\rm{\pi }}} \cos x \ dx\)

= \(\rm √ 2 \left[ {\sin x} \right]_0^{{{\rm{2\pi }}}}\)

= \(\rm √ 2 [\sin {2\pi\ -\ \sin 0}]\)

= 0