<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>
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Correct Answer: D