Evaluate the integrals 1−cos⁡x2dx

Consider the given integral ${\displaystyle {\int }_0^{2\pi }\sqrt{\frac{1-\cos x}{2}}dx}$
Use the property of ${\displaystyle \cos x=1-2{\sin }^2\frac{x}{2}}$
${\displaystyle {\int }_0^{2\pi }\sqrt{\frac{1-\cos x}{2}}dx={\int }_0^{2\pi }\sqrt{\frac{1-(1-2{\sin }^2\frac{x}{2}\}\left\{2\right\}}{dx}}}$
${\displaystyle ={\int }_0^{2\pi }\sqrt{\frac{2{\sin }^2\frac{x}{2}}{2}}dx}$
${\displaystyle ={\int }_0^{2\pi }\sqrt{\frac{2{\sin }^2\frac{x}{2}}{2}}dx}$
${\displaystyle ={\int }_0^{2\pi }\sqrt{\frac{2{\sin }^2\frac{x}{2}}{2}}dx}$
${\displaystyle ={\int }_0^{2\pi }\sqrt{{\sin }^2\frac{x}{2}}dx}$
As we know that ${\displaystyle \sqrt{a^2}=\left|a\right|}$
${\displaystyle {\int }_0^{2\pi }\sqrt{\frac{1-\cos x}{2}}dx={\int }_0^{2\pi }\left|\sin \frac{x}{2}\right|dx}$
But ${\displaystyle \left|\sin \frac{x}{2}\right|=\sin \frac{x}{2}}$ because interval given $(0,2\pi )$
${\displaystyle {\int }_0^{2\pi }\sqrt{\frac{1-\cos x}{2}}dx={\int }_0^{2\pi }\sin \frac{x}{2}dx}$
${\displaystyle ={\left[\frac{(-\cos \frac{x}{2})}{\frac{1}{2}}\right]}_0^{2\pi }}$
${\displaystyle =-2(\cos \frac{2\pi }{2}=\cos 0)}$
${\displaystyle =-2(\cos \pi -\cos 0)}$