Question

The anti-derivative of \( \sqrt{1 + \sin 2x}, \, x \in \left[ 0, \frac{\pi}{4} \right] \) is:

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

Hint:

To solve integrals involving trigonometric expressions, use trigonometric identities like the double-angle formula to simplify the expression before integration.

Answer 1

The Correct Option is B

Step 1: {Rewrite the given integral}
The problem is to find the anti-derivative of \( \sqrt{1 + \sin 2x} \). Using the trigonometric identity: \[ \sin 2x = 2\sin x \cos x, \] we rewrite \( \sqrt{1 + \sin 2x} \) as: \[ \sqrt{1 + 2\sin x \cos x}. \] Step 2: {Simplify using the double-angle formula}
The expression \( 1 + 2\sin x \cos x \) can be written as: \[ 1 + \sin 2x = (\sin x + \cos x)^2. \] Thus: \[ \sqrt{1 + \sin 2x} = |\sin x + \cos x|. \] Step 3: {Evaluate the integral}
For \( x \in \left[ 0, \frac{\pi}{4} \right] \), both \( \sin x \) and \( \cos x \) are positive, so: \[ \sqrt{1 + \sin 2x} = \sin x + \cos x. \] The anti-derivative is: \[ \int (\sin x + \cos x) \, dx = -\cos x + \sin x + C. \] Conclusion: The anti-derivative of \( \sqrt{1 + \sin 2x} \) is \( -\cos x + \sin x \).