Question

\(\int_{0}^{\pi/2} \frac{\cos 2x}{\sin x + \cos x} \,dx =\)

• A. -1
• B. 0
• C. 1
• D. \(\frac{\pi}{2}\)

Hint:

When faced with a trigonometric fraction in an integral, always look for identities that can simplify the expression. The double angle formulas, especially \(\cos 2x\), are very versatile and have multiple forms (\(\cos^2x - \sin^2x\), \(2\cos^2x - 1\), \(1 - 2\sin^2x\)). Choosing the right form is key. Here, the difference of squares form was perfect for cancellation.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to evaluate a definite integral involving trigonometric functions.

Step 2: Key Formula or Approach:
The integrand looks complicated. We should try to simplify it using trigonometric identities. The key is to relate the numerator \(\cos 2x\) to the denominator \(\sin x + \cos x\).
The relevant identity for the numerator is the difference of squares form:
\[ \cos 2x = \cos^2 x - \sin^2 x \]
This can be factored as \((\cos x - \sin x)(\cos x + \sin x)\).

Step 3: Detailed Explanation:
Let \(I = \int_{0}^{\pi/2} \frac{\cos 2x}{\sin x + \cos x} \,dx\).
Substitute the identity for \(\cos 2x\) into the integral:
\[ I = \int_{0}^{\pi/2} \frac{\cos^2 x - \sin^2 x}{\sin x + \cos x} \,dx \]
Factor the numerator as a difference of squares:
\[ I = \int_{0}^{\pi/2} \frac{(\cos x - \sin x)(\cos x + \sin x)}{\sin x + \cos x} \,dx \]
Assuming \(\sin x + \cos x \neq 0\) in the interval \((0, \pi/2)\), we can cancel the common term:
\[ I = \int_{0}^{\pi/2} (\cos x - \sin x) \,dx \]
Now, we can integrate this simplified expression:
\[ I = \left[ \sin x - (-\cos x) \right]_{0}^{\pi/2} \]
\[ I = \left[ \sin x + \cos x \right]_{0}^{\pi/2} \]
Evaluate 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 definite integral is 0.