integration under root (sin2x -1)

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{I} \mathrm{think} \mathrm{it} \mathrm{should} \mathrm{be} \sqrt{1-\sin 2\mathrm{x}} \mathrm{because} \mathrm{as} -1\le \sin 2\mathrm{x}\le 1, \mathrm{so} \sqrt{\sin 2\mathrm{x}-1} \mathrm{will} \mathrm{only} \mathrm{be} \mathrm{defined} \mathrm{when} \\ \sin 2\mathrm{x}=1, \mathrm{so} \mathrm{f}\left(\mathrm{x}\right)=\sqrt{\sin 2\mathrm{x}-1} \mathrm{will} \mathrm{always} \mathrm{be} 0. \left[\sqrt{1-1}=0\right] \\ \mathrm{So}, \mathrm{we} \mathrm{have}, \\ \mathrm{I}=\int \sqrt{1-\sin 2\mathrm{x}}\mathrm{dx} \\ =\int \sqrt{{\sin }^2+{\cos }^2\mathrm{x}-2\mathrm{sinxcosx}}\mathrm{dx} \\ =\int \sqrt{{\left(\mathrm{sinx}-\mathrm{cosx}\right)}^2}\mathrm{dx} \\ =\int \left(\mathrm{sinx}-\mathrm{cosx}\right)\mathrm{dx} \\ =-\mathrm{cosx}-\mathrm{sinx} +\mathrm{C} \\ \mathbf{\Rightarrow }\mathbf{\int }\sqrt{\mathbf{1}\mathbf{-}\mathbf{sin}\mathbf{2}\mathbf{x}}\mathbf{dx}\mathbf{=}\mathbf{-}\left(\mathbf{sinx}\mathbf{+}\mathbf{cosx}\right)\mathbf{+}\mathbf{C} \end{gathered}$$

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