Correct Answer: A $(0,1)$ The solution to this integral can be found by recognizing that the integrand is the derivative of a simpler function. This allows direct evaluation using the Fundamental Theorem of Calculus. Step 1: Identify the Antiderivative Consider the function \[ f(x) = \frac{\sin x}{2+\cos x} \] Using the quotient rule, \[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \] Let \( u = \sin x \) and \( v = 2+\cos x \). Then \( u' = \cos x \) and \( v' = -\sin x \). \[ f'(x) = \frac{(\cos x)(2+\cos x) - (\sin x)(-\sin x)}{(2+\cos x)^2} \] \[ = \frac{2\cos x + \cos^2 x + \sin^2 x}{(2+\cos x)^2} \] Since \( \sin^2 x + \cos^2 x = 1 \), \[ f'(x) = \frac{1 + 2\cos x}{(2+\cos x)^2} \] Step 2: Evaluate the Definite Integral \[ I = \int_{0}^{\frac{\pi}{2}} \frac{1+2\cos x}{(2+\cos x)^2} \, dx \] Since the integrand is \( f'(x) \), \[ I = \left[ \frac{\sin x}{2+\cos x} \right]_{0}^{\frac{\pi}{2}} \] Upper limit \( x = \frac{\pi}{2} \): \[ \frac{\sin\left(\frac{\pi}{2}\right)}{2+\cos\left(\frac{\pi}{2}\right)} = \frac{1}{2} \] Lower limit \( x = 0 \): \[ \frac{\sin(0)}{2+\cos(0)} = 0 \] \[ I = \frac{1}{2} - 0 = \frac{1}{2} \] Step 3: Determine the Interval \[ I = \frac{1}{2} \] Since \( 0 < \frac{1}{2} < 1 \), the value lies in \[ (0,1) \]