Question

If \( x \notin \left[ 2n\pi - \frac{\pi}{4}, 2n\pi + \frac{3\pi}{4} \right] \) and \( n \in \mathbb{Z} \), then \[ \int \sqrt{1 - \sin 2x} \, dx = \]

• A. \( -\cos x + \sin x + c \)
• B. \( \cos x + \sin x + c \)
• C. \( \cos x - \sin x + c \)
• D. \( -\cos x - \sin x + c \)

Hint:

For integrals involving \( \sqrt{1 - \sin 2x} \), express it as \( (\cos x - \sin x)^2 \) and simplify using absolute value properties.

Answer 1

The Correct Option is B

Step 1: Substituting the given integral 
We need to evaluate: \[ I = \int \sqrt{1 - \sin 2x} \, dx. \] Using the identity: \[ 1 - \sin 2x = \cos^2 x + \sin^2 x - \sin 2x. \] Using the identity: \[ 1 - \sin 2x = (\cos x - \sin x)^2. \] Thus, \[ I = \int \sqrt{(\cos x - \sin x)^2} \, dx. \] Since \( \sqrt{(\cos x - \sin x)^2} = |\cos x - \sin x| \), we need to determine its sign.
Step 2: Evaluating \( \cos x - \sin x \) 
- If \( x \notin \left[ 2n\pi - \frac{\pi}{4}, 2n\pi + \frac{3\pi}{4} \right] \), then \( \cos x - \sin x \) is positive. 
- Thus, \( |\cos x - \sin x| = \cos x - \sin x \). \[ I = \int (\cos x - \sin x) \, dx. \] Step 3: Evaluating the integral 
\[ I = \int \cos x \, dx - \int \sin x \, dx. \] \[ I = \sin x + \cos x + c. \] Step 4: Conclusion 
Thus, the correct answer is: \[ \cos x + \sin x + c. \]