If $x\in \left(\pi ,\frac{{\displaystyle 3\pi }}{{\displaystyle 2}}\right)$then $\int (\sqrt{1+\sin x}-\sqrt{1-\sin x})dx=$

$$\begin{gathered} \mathrm{x}\in \left(\mathrm{\pi },\frac{{\displaystyle 3\mathrm{\pi }}}{{\displaystyle 2}}\right) \\ \frac{\mathrm{x}}{2}\in \left(\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{4}\right) \\ \sqrt{1+\sin \mathrm{x}}-\sqrt{1-\sin \mathrm{x}} \\ =\left|\sin \frac{\mathrm{x}}{2}+\cos \frac{\mathrm{x}}{2}\right|-\left|\sin \frac{\mathrm{x}}{2}-\cos \frac{x}{2}\right| \\ =\sin \frac{\mathrm{x}}{2}+\cos \frac{\mathrm{x}}{2}-\left(\sin \frac{\mathrm{x}}{2}-\cos \frac{\mathrm{x}}{2}\right) \\ \mathrm{I}=\int 2\cos \frac{\mathrm{x}}{2}\mathrm{dx} \\ =2 \frac{\sin {\displaystyle \frac{\mathrm{x}}{2}}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=4\sin \frac{\mathrm{x}}{2}+\mathrm{C} \end{gathered}$$