Question

$\int_{0}^{1}\frac{\sin x}{\sin x+\sin(1-x)}\,dx$ is equal to:

• A. \(6\)
• B. \(4\)
• C. \(2\)
• D. \(\frac{1}{2}\)
• E. \(\frac{1}{4}\)

Hint:

For definite integrals involving \(f(x)\) and \(f(a-x)\), always try substitution \(x \to a-x\) and add both expressions.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The problem shows an indefinite integral, but the options are all constants. This is a strong indication that it's a definite integral where the limits of integration have been omitted in the transcription. The structure of the integrand, \( f(x) / (f(x) + f(a+b-x)) \), points towards a standard property of definite integrals, often with limits from \( a \) to \( b \). Given the term \( (1-x) \), the most likely limits are from 0 to 1.
Step 2: Key Formula or Approach:
We will assume the question is to evaluate \( I = \int_0^1 \frac{\sin x}{\sin x + \sin(1-x)} dx \).
We use the property of definite integrals:
\[ \int_a^b f(x) dx = \int_a^b f(a+b-x) dx \] Step 3: Detailed Explanation:
Let the integral be \( I = \int_0^1 \frac{\sin x}{\sin x + \sin(1-x)} dx \). This is our equation (1).
Using the property \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \), with \( a=1 \), we replace \( x \) with \( (1-x) \):
\[ I = \int_0^1 \frac{\sin(1-x)}{\sin(1-x) + \sin(1-(1-x))} dx \] \[ I = \int_0^1 \frac{\sin(1-x)}{\sin(1-x) + \sin x} dx \] This is our equation (2).
Now, we add equation (1) and equation (2):
\[ I + I = \int_0^1 \frac{\sin x}{\sin x + \sin(1-x)} dx + \int_0^1 \frac{\sin(1-x)}{\sin(1-x) + \sin x} dx \] \[ 2I = \int_0^1 \frac{\sin x + \sin(1-x)}{\sin x + \sin(1-x)} dx \] Since the numerator and denominator are identical, the integrand simplifies to 1:
\[ 2I = \int_0^1 1 dx \] Evaluating this simple integral:
\[ 2I = [x]_0^1 = 1 - 0 = 1 \] \[ I = \frac{1}{2} \] Step 4: Final Answer:
Assuming the intended question was a definite integral from 0 to 1, the value is \( \frac{1}{2} \). This corresponds to option (D).