The integral `int_(0)^(pi) x f(sinx )dx` is equal to

Options

A. `(pi)/(2)underset(0)overset(pi)intf(sin x)dx`
B. `(pi)/(4)underset(0)overset(pi)int(sin x)dx`
C. `(pi)/(2)underset(0)overset(pi//2)int (sin x)dx`
D. `(pi)/(2)underset(0)overset(pi//2)intf(cos x)dx`

Explanation

To solve the integral \( I = \int_{0}^{\pi} x f(\sin x) \, dx \), we will use the property of definite integrals which states that:

\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]

In our case, \( a = \pi \). Let's proceed step by step.

### Step 1: Set up the integral
Let:
\[ I = \int_{0}^{\pi} x f(\sin x) \, dx \]

### Step 2: Apply the property of definite integrals
Using the property mentioned, we can rewrite the integral as follows:
\[ I = \int_{0}^{\pi} (\pi - x) f(\sin(\pi - x)) \, dx \]

### Step 3: Simplify the sine function
Since \( \sin(\pi - x) = \sin x \), we can substitute this into the integral:
\[ I = \int_{0}^{\pi} (\pi - x) f(\sin x) \, dx \]

### Step 4: Expand the integral
Now, we can expand the integral:
\[ I = \int_{0}^{\pi} \pi f(\sin x) \, dx - \int_{0}^{\pi} x f(\sin x) \, dx \]

### Step 5: Recognize the original integral
Notice that the second term on the right side is just \( I \):
\[ I = \pi \int_{0}^{\pi} f(\sin x) \, dx - I \]

### Step 6: Combine like terms
Now, we can add \( I \) to both sides:
\[ 2I = \pi \int_{0}^{\pi} f(\sin x) \, dx \]

### Step 7: Solve for \( I \)
Dividing both sides by 2 gives us:
\[ I = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \, dx \]

### Final Result
Thus, the value of the integral \( \int_{0}^{\pi} x f(\sin x) \, dx \) is:
\[ \boxed{\frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \, dx} \]

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