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 can use the property of definite integrals that states:

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

In this case, \( a = 0 \) and \( b = \pi \). Let's apply this property step by step.

### Step 1: Rewrite the integral using the property
We can express \( I \) as follows:

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

### Step 2: Simplify the function inside the integral
Using the identity \( \sin(\pi - x) = \sin x \), we can rewrite the integral:

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

### Step 3: Split the integral
Now we can split the integral into two parts:

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

### Step 4: Combine the integrals
Notice that the second integral on the right side is just \( I \):

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

### Step 5: Solve for \( I \)
Now, we can add \( I \) to both sides:

\[ 2I = \pi \int_{0}^{\pi} f(\sin x) \, dx \]

Now, divide both sides by 2:

\[ I = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \, dx \]

### Final Result
Thus, the value of the integral is:

\[ \int_{0}^{\pi} x f(\sin x) \, dx = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \, dx \]

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