Question

The value of \( \displaystyle \int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x} \, dx \) is:

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

Hint:

For definite integrals involving \( \sin x \) and \( \cos x \), try the substitution \( x \rightarrow a-x \). This often converts the integral into a complementary form that simplifies when added.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem asks for the evaluation of a definite integral involving a trigonometric ratio over the interval \( [0, \pi/2] \).
The topic is Definite Integration, specifically focusing on the properties of symmetric functions.
Step 2: Key Formula or Approach:
We use the property of definite integrals:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
Applying this to the given limits, we replace \( x \) with \( \frac{\pi}{2} - x \).
Step 3: Detailed Explanation:
Let the given integral be \( I \):
\[ I = \int_{0}^{\pi/2} \frac{\sin x}{\sin x + \cos x} \, dx \quad \text{--- (i)} \]
Using the property \( x \to \frac{\pi}{2} - x \):
\[ I = \int_{0}^{\pi/2} \frac{\sin(\pi/2 - x)}{\sin(\pi/2 - x) + \cos(\pi/2 - x)} \, dx \]
Using the identities \( \sin(\pi/2 - x) = \cos x \) and \( \cos(\pi/2 - x) = \sin x \):
\[ I = \int_{0}^{\pi/2} \frac{\cos x}{\cos x + \sin x} \, dx \quad \text{--- (ii)} \]
Adding equations (i) and (ii):
\[ 2I = \int_{0}^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x} \, dx \]
\[ 2I = \int_{0}^{\pi/2} 1 \, dx \]
\[ 2I = [x]_{0}^{\pi/2} \]
\[ 2I = \frac{\pi}{2} - 0 \]
Step 4: Final Answer:
Solving for \( I \):
\[ I = \frac{\pi}{4} \]