Let `I_(1) =int_(a)^(pi-a)xf(sinx)dx,I_(2)=int_(a)^(pi-a)f(sinx)dx`, then `I_(2)` is equal to

Options

A. `(pi)/2 I_(1)`
B. `piI_(1)`
C. `2/(pi)I_(1)`
D. `2I_(1)`

Answer

To solve the problem, we need to find the value of \( I_2 = \int_a^{\pi - a} f(\sin x) \, dx \) given that \( I_1 = \int_a^{\pi - a} x f(\sin x) \, dx \).

### Step-by-step solution:

1. **Understanding the Integrals**:
We have two integrals:
\[ I_1 = \int_a^{\pi - a} x f(\sin x) \, dx \]
\[ I_2 = \int_a^{\pi - a} f(\sin x) \, dx \]

2. **Using the Property of Definite Integrals**:
We can use the property of definite integrals that states:
\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]
In our case, we can apply this property to \( I_1 \).

3. **Changing the Variable in \( I_1 \)**:
Let’s change the variable in \( I_1 \) using \( x = \pi - a - t \). Then, when \( x = a \), \( t = \pi - 2a \) and when \( x = \pi - a \), \( t = 0 \). The differential \( dx = -dt \).

Thus, we have:
\[ I_1 = \int_{a}^{\pi - a} x f(\sin x) \, dx = \int_{\pi - a}^{a} (\pi - a - t) f(\sin(\pi - a - t)) (-dt) \]
Since \( \sin(\pi - x) = \sin x \), we can simplify this to:
\[ I_1 = \int_{a}^{\pi - a} (\pi - a - t) f(\sin t) \, dt \]

4. **Splitting the Integral**:
Now, we can split the integral:
\[ I_1 = \int_a^{\pi - a} (\pi - a) f(\sin t) \, dt - \int_a^{\pi - a} t f(\sin t) \, dt \]
This gives us:
\[ I_1 = (\pi - a) I_2 - I_1 \]

5. **Solving for \( I_2 \)**:
Rearranging the equation gives:
\[ 2I_1 = (\pi - a) I_2 \]
Therefore:
\[ I_2 = \frac{2I_1}{\pi - a} \]

### Final Result:
Thus, we have:
\[ I_2 = \frac{2I_1}{\pi - a} \]