Let f(x) be a function defined on R satisfyin `f(x) =f(1-x)` for all ` x in R`. Then `int_(-1//2)^(1//2) f(x+(1)/(2))sin x dx` equals

Options

A. `underset(-1//2)overset(1//2)int f(x+(1)/(2))sin x dx`
B. `2underset(-1//2)overset(1//2)int f(x+(1)/(2))sin x dx`
C. `f(x) sind x dx`
D. none of these

Explanation

To solve the given integral \( I = \int_{-\frac{1}{2}}^{\frac{1}{2}} f\left(x + \frac{1}{2}\right) \sin x \, dx \), where \( f(x) = f(1 - x) \), we will follow these steps:

### Step 1: Change of Variable
Let's first perform a change of variable in the integral. We will let \( u = x + \frac{1}{2} \). Then, we have:
\[ du = dx \quad \text{and} \quad x = u - \frac{1}{2} \]
The limits of integration change as follows:
- When \( x = -\frac{1}{2} \), \( u = 0 \).
- When \( x = \frac{1}{2} \), \( u = 1 \).

Thus, the integral becomes:
\[ I = \int_{0}^{1} f(u) \sin\left(u - \frac{1}{2}\right) \, du \]

### Step 2: Use the Property of \( f(x) \)
Using the property \( f(x) = f(1 - x) \), we can express \( f(u) \) in terms of \( f(1 - u) \):
\[ I = \int_{0}^{1} f(1 - u) \sin\left(u - \frac{1}{2}\right) \, du \]

### Step 3: Change of Variable in the Integral
Now, we will change the variable again. Let \( v = 1 - u \). Then, \( du = -dv \) and the limits change as follows:
- When \( u = 0 \), \( v = 1 \).
- When \( u = 1 \), \( v = 0 \).

Thus, the integral becomes:
\[ I = \int_{1}^{0} f(v) \sin\left(1 - v - \frac{1}{2}\right)(-dv) = \int_{0}^{1} f(v) \sin\left(\frac{1}{2} - v\right) \, dv \]

### Step 4: Combine the Integrals
Now we have two expressions for \( I \):
1. \( I = \int_{0}^{1} f(u) \sin\left(u - \frac{1}{2}\right) \, du \)
2. \( I = \int_{0}^{1} f(v) \sin\left(\frac{1}{2} - v\right) \, dv \)

### Step 5: Add the Two Integrals
Adding these two integrals:
\[ 2I = \int_{0}^{1} f(u) \left(\sin\left(u - \frac{1}{2}\right) + \sin\left(\frac{1}{2} - u\right)\right) \, du \]

Using the identity \( \sin(a) + \sin(b) = 2 \sin\left(\frac{a+b}{2}\right) \cos\left(\frac{a-b}{2}\right) \):
\[ \sin\left(u - \frac{1}{2}\right) + \sin\left(\frac{1}{2} - u\right) = 2 \sin\left(0\right) \cos\left(u - \frac{1}{2}\right) = 0 \]

Thus:
\[ 2I = 0 \implies I = 0 \]

### Conclusion
Therefore, we conclude that:
\[ \int_{-\frac{1}{2}}^{\frac{1}{2}} f\left(x + \frac{1}{2}\right) \sin x \, dx = 0 \]