If \( 0\lt x\lt\displaystyle\frac{\pi}{4} \), then \( \int\displaystyle\sqrt{1-\sin 2x}\,dx \) is:
[ where C is an arbitrary constant]
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \sin x+\cos x+C \)
Since \( 1=\sin^2x+\cos^2x \) and \( \sin 2x=2\sin x\cos x \), the expression under the root is a perfect square:
\( \displaystyle 1-\sin 2x=\sin^2x+\cos^2x-2\sin x\cos x \)
\( \displaystyle =(\cos x-\sin x)^2 \)
\( \displaystyle \sqrt{1-\sin 2x}=|\cos x-\sin x| \)
For \( 0\lt x\lt\frac{\pi}{4} \), cos x > sin x, so \( |\cos x-\sin x|=\cos x-\sin x \).
\( \displaystyle \int(\cos x-\sin x)\,dx=\sin x+\cos x+C \)
Why the other options are incorrect
Option (2) \( \sin x-\cos x+C \) — Incorrect. Its derivative is \( \cos x+\sin x \), not \( \cos x-\sin x \).
Option (3) \( -\sin x-\cos x+C \) — Incorrect. This is the integral of \( \sin x-\cos x \), which is negative on \( \left(0,\frac{\pi}{4}\right) \); the root must be taken as \( \cos x-\sin x \) because a square root is never negative.
Option (4) \( -\sin x+\cos x+C \) — Incorrect. Its derivative is \( -\cos x-\sin x \), which is negative, while \( \sqrt{1-\sin 2x}\gt 0 \).