<p>Evaluate \(\displaystyle\int_0^{\pi/2}\sqrt{1-\sin 2x}\,dx\) [JEE Main 2020]</p>
Step-by-Step Solution
Key Concept: 1 - sin2x = (sinx - cosx)^2. So \sqrt{1-sin2x} = |sinx - cosx|. Split at x=\pi/4.
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<p>$1-\sin 2x=\sin^2 x-2\sin x\cos x+\cos^2 x=(\sin x-\cos x)^2$</p>
<p>$\sqrt{1-\sin 2x}=|\sin x-\cos x|$</p>
<p>On $[0,\pi/4]$: $\cos x>\sin x$, so $|\sin x-\cos x|=\cos x-\sin x$.</p>
<p>On $[\pi/4,\pi/2]$: $\sin x>\cos x$, so $|\sin x-\cos x|=\sin x-\cos x$.</p>
<p>$$I=\int_0^{\pi/4}(\cos x-\sin x)dx+\int_{\pi/4}^{\pi/2}(\sin x-\cos x)dx$$</p>
<p>$=[\sin x+\cos x]_0^{\pi/4}+[-\cos x-\sin x]_{\pi/4}^{\pi/2}$</p>
<p>$=(\sqrt{2}-1)+(-1-(-\sqrt{2}))=(\sqrt{2}-1)+(\sqrt{2}-1)=2(\sqrt{2}-1)$</p>
Correct Answer: D