If `int_(0)^(pi) x f(sin x)dx= k int_(0)^(pi//2) f(sin x) dx` then the value of k is

Options

A. 2
B. 1
C. `pi`
D. 0

Answer

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 \( \int_{0}^{\frac{\pi}{2}} f(\sin x) \, dx \) to find the value of \( k \).

### Step-by-Step Solution:

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

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

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

4. **Expand the Integral**:
Now we can expand this integral:
\[ I = \int_{0}^{\pi} \pi f(\sin x) \, dx - \int_{0}^{\pi} x f(\sin x) \, dx \]
This gives us:
\[ I = \pi \int_{0}^{\pi} f(\sin x) \, dx - I \]

5. **Combine Like Terms**:
Adding \( I \) to both sides results in:
\[ 2I = \pi \int_{0}^{\pi} f(\sin x) \, dx \]

6. **Solve for \( I \)**:
Dividing both sides by 2 gives:
\[ I = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \, dx \]

7. **Relate to the Given Equation**:
We know from the problem statement:
\[ I = k \int_{0}^{\frac{\pi}{2}} f(\sin x) \, dx \]

8. **Use the Property of Integrals Again**:
We can relate the integral from \( 0 \) to \( \pi \) to the integral from \( 0 \) to \( \frac{\pi}{2} \):
\[ \int_{0}^{\pi} f(\sin x) \, dx = 2 \int_{0}^{\frac{\pi}{2}} f(\sin x) \, dx \]
Hence,
\[ I = \frac{\pi}{2} \cdot 2 \int_{0}^{\frac{\pi}{2}} f(\sin x) \, dx = \pi \int_{0}^{\frac{\pi}{2}} f(\sin x) \, dx \]

9. **Compare Coefficients**:
From the equations \( I = k \int_{0}^{\frac{\pi}{2}} f(\sin x) \, dx \) and \( I = \pi \int_{0}^{\frac{\pi}{2}} f(\sin x) \, dx \), we can equate:
\[ k = \pi \]

### Final Result:
Thus, the value of \( k \) is \( \pi \).