∫_0^π/2√(1-sin 2x) dx is equal to:

The correct answer is Option (4) → \(2(\sqrt{2}-1)\)

Since \(1=\sin^2x+\cos^2x\) and \(\sin 2x=2\sin x\cos x\):

    \( 1-\sin 2x=\sin^2x+\cos^2x-2\sin x\cos x \)

    \( =(\cos x-\sin x)^2 \)

So

    \( \sqrt{1-\sin 2x}=|\cos x-\sin x| \)

The sign changes at \(x=\frac{\pi}{4}\): \(\cos x\ge\sin x\) on \([0, \frac{\pi}{4}]\) and \(\sin x\ge\cos x\) on \([\frac{\pi}{4}, \frac{\pi}{2}]\). Splitting the integral there:

    \( \displaystyle I=\int_{0}^{\pi/4}(\cos x-\sin x)\,dx+\int_{\pi/4}^{\pi/2}(\sin x-\cos x)\,dx \)

    \( \displaystyle =\Big[\sin x+\cos x\Big]_{0}^{\pi/4}+\Big[-\cos x-\sin x\Big]_{\pi/4}^{\pi/2} \)

    \( =(\sqrt{2}-1)+(-1+\sqrt{2}) \)

    \( =2(\sqrt{2}-1) \)

Tip: \(\sqrt{(\cos x-\sin x)^2}\) is \(|\cos x-\sin x|\), not \(\cos x-\sin x\). Dropping the modulus here gives 0 for the whole integral.