Question

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

• A. \( \dfrac{\pi}{2} \)
• B. \( \dfrac{\pi}{3} \)
• C. \( \dfrac{\pi}{4} \)
• D. \( \dfrac{\pi}{6} \)

Hint:

When a definite integral contains symmetric expressions like \(\sin x\) and \(\cos x\), applying the property \[ \int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx \] often simplifies the integral significantly.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to evaluate a definite integral with limits from \(0\) to \(\pi/2\).
The integrand involves trigonometric functions that exhibit symmetry around the midpoint of the interval.
Step 2: Key Formula or Approach:
Use the "King's Property" of definite integrals:
\[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \]
Step 3: Detailed Explanation:
Let \( I = \int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x} \, dx \) --- (Eq. 1)
Applying the property \(\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx\):
\[ I = \int_{0}^{\pi/2} \frac{\sin(\pi/2 - x)}{\sin(\pi/2 - x) + \cos(\pi/2 - x)} \, dx \]
Since \(\sin(\pi/2 - x) = \cos x\) and \(\cos(\pi/2 - x) = \sin x\):
\[ I = \int_{0}^{\pi/2} \frac{\cos x}{\cos x + \sin x} \, dx \] --- (Eq. 2)
Adding (Eq. 1) and (Eq. 2):
\[ 2I = \int_{0}^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x} \, dx \]
\[ 2I = \int_{0}^{\pi/2} 1 \, dx \]
\[ 2I = [x]_{0}^{\pi/2} = \frac{\pi}{2} \]
\[ I = \frac{\pi}{4} \]
Step 4: Final Answer:
The value of the integral is \( \dfrac{\pi}{4} \).