If `int_(0)^(pi) x f (sin x) dx = A int _(0)^(pi//2) f(sin x) dx ,` then A is equal to

Options

A. 0
B. `pi`
C. `(pi)/(4)`
D. `2pi`

Explanation

To solve the problem, we need to evaluate the integral \( I = \int_{0}^{\pi} x f(\sin x) \, dx \) and relate it to the integral \( \int_{0}^{\frac{\pi}{2}} f(\sin x) \, dx \).

### Step-by-Step Solution:

1. **Define the Integral**:
Let \( I = \int_{0}^{\pi} x f(\sin x) \, dx \).

2. **Use the Property of Definite Integrals**:
We can use the property of definite integrals that states:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
In our case, set \( a = \pi \). Thus, we have:
\[ I = \int_{0}^{\pi} x f(\sin x) \, dx = \int_{0}^{\pi} (\pi - x) f(\sin(\pi - x)) \, dx \]

3. **Simplify the Integral**:
Since \( \sin(\pi - x) = \sin x \), we can rewrite the integral as:
\[ I = \int_{0}^{\pi} (\pi - x) f(\sin x) \, dx \]

4. **Combine the Integrals**:
Now we have two expressions for \( I \):
\[ I = \int_{0}^{\pi} x f(\sin x) \, dx \]
and
\[ I = \int_{0}^{\pi} (\pi - x) f(\sin x) \, dx \]
Adding these two equations gives:
\[ 2I = \int_{0}^{\pi} \left( x + (\pi - x) \right) f(\sin x) \, dx = \int_{0}^{\pi} \pi f(\sin x) \, dx \]

5. **Factor Out Constants**:
Thus, we can simplify:
\[ 2I = \pi \int_{0}^{\pi} f(\sin x) \, dx \]
Therefore, we find:
\[ I = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \, dx \]

6. **Relate to the Given Equation**:
The problem states:
\[ I = A \int_{0}^{\frac{\pi}{2}} f(\sin x) \, dx \]
We need to express \( \int_{0}^{\pi} f(\sin x) \, dx \) in terms of \( \int_{0}^{\frac{\pi}{2}} f(\sin x) \, dx \).

7. **Use Symmetry**:
The integral \( \int_{0}^{\pi} f(\sin x) \, dx \) can be split into two equal parts due to the symmetry of the sine function:
\[ \int_{0}^{\pi} f(\sin x) \, dx = 2 \int_{0}^{\frac{\pi}{2}} f(\sin x) \, dx \]

8. **Substitute Back**:
Substituting this back into our expression for \( I \):
\[ I = \frac{\pi}{2} \cdot 2 \int_{0}^{\frac{\pi}{2}} f(\sin x) \, dx = \pi \int_{0}^{\frac{\pi}{2}} f(\sin x) \, dx \]

9. **Identify A**:
Comparing this with the original equation \( I = A \int_{0}^{\frac{\pi}{2}} f(\sin x) \, dx \), we find:
\[ A = \pi \]

### Final Answer:
Thus, the value of \( A \) is \( \pi \).