 ediculeN

2021-10-06

Evaluate the integrals
$\sqrt{\frac{1-\mathrm{cos}x}{2}}dx$ Raheem Donnelly

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

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