Question

$\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\sqrt{1-\sin 2x}dx$ is equal to

(a) $2\sqrt{2}$

(b) $2\left(\sqrt{2}+1\right)$

(c) 2

(b) $2\left(\sqrt{2}-1\right)$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\sqrt{1-\sin 2x}dx \\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\sqrt{{\cos }^{2}x+{\sin }^{2}x-2\sin x\cos x}dx\left(\mathrm{using}\mathrm{the}\mathrm{identity}:{\cos }^{2}x+{\sin }^{2}x=1\mathrm{and}2\sin x\cos x=\sin 2x\right) \\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\sqrt{{\left(\cos x-\sin x\right)}^2}dx \\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\left|\cos x-\sin x\right|dx \\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}\left|\cos x-\sin x\right|dx+\underset{\frac{\mathrm{\pi }}{4}}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\left|\cos x-\sin x\right|dx \\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}\left(\cos x-\sin x\right)dx+\underset{\frac{\mathrm{\pi }}{4}}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}-\left(\cos x-\sin x\right)dx \\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}\left(\cos x-\sin x\right)dx+\underset{\frac{\mathrm{\pi }}{4}}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\left(\sin x-\cos x\right)dx \\ ={\left[\sin x+\cos x\right]}_0^{\frac{\mathrm{\pi }}{4}}+{\left[-\cos x-\sin x\right]}_{\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{2}} \\ =\left[\sin \frac{\mathrm{\pi }}{4}+\cos \frac{\mathrm{\pi }}{4}-\sin 0-\cos 0\right]+\left[-\cos \frac{\mathrm{\pi }}{2}-\sin \frac{\mathrm{\pi }}{2}-\left(-\cos \frac{\mathrm{\pi }}{4}-\sin \frac{\mathrm{\pi }}{4}\right)\right] \\ =\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-0-1\right]+\left[-0-1-\left(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right)\right] \\ =\left[\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1\right]+\left[-1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right] \\ =\frac{2}{\sqrt{2}}-1+-1+\frac{2}{\sqrt{2}} \\ =\sqrt{2}-1+-1+\sqrt{2} \\ =2\sqrt{2}-2 \\ =2\left(\sqrt{2}-1\right) \end{gathered}$$


Hence, the correct option is (d).