Step 1: Understanding the Concept:
This definite integral can be simplified by using a trigonometric identity for the numerator that relates to the denominator, allowing for cancellation.
Step 2: Key Formula or Approach:
The key is to use the double-angle identity for cosine in the form of a difference of squares:
\[ \cos(2x) = \cos^2 x - \sin^2 x \]
This can be factored as:
\[ \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x) \]
Step 3: Detailed Explanation:
The integral is $I = \int_0^{\pi/2} \frac{\cos 2x}{\sin x + \cos x} \,dx$.
Substitute the identity for $\cos(2x)$ in the numerator:
\[ I = \int_0^{\pi/2} \frac{\cos^2 x - \sin^2 x}{\sin x + \cos x} \,dx \]
Factor the numerator as a difference of squares:
\[ I = \int_0^{\pi/2} \frac{(\cos x - \sin x)(\cos x + \sin x)}{\sin x + \cos x} \,dx \]
Since $\sin x + \cos x$ is not zero in the interval $(0, \pi/2)$, we can cancel the term:
\[ I = \int_0^{\pi/2} (\cos x - \sin x) \,dx \]
Now, integrate the simplified expression:
\[ I = \left[ \sin x - (-\cos x) \right]_0^{\pi/2} = \left[ \sin x + \cos x \right]_0^{\pi/2} \]
Evaluate the integral at the upper and lower limits:
\[ I = \left(\sin\left(\frac{\pi}{2}\right) + \cos\left(\frac{\pi}{2}\right)\right) - (\sin(0) + \cos(0)) \]
\[ I = (1 + 0) - (0 + 1) \]
\[ I = 1 - 1 = 0 \]
Step 4: Final Answer:
The value of the integral is 0. Therefore, option (B) is correct.