`int_(0)^(pi) x f (sin x)dx`=

Options

A. `pi int_(0)^(pi) f (sin x) dx`
B. `(pi)/(2) int_(0)^(pi//2) f(sin x)dx`
C. `pi int_(0)^(pi//2) f (cos x)dx`
D. `pi int_(0)^(pi) f (cos x) dx`

Answer

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

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

### Step 2: Use the property of definite integrals
We can use the property that states:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
In our case, we will apply this property with \( a = \pi \):
\[ I = \int_{0}^{\pi} x f(\sin x) \, dx = \int_{0}^{\pi} (\pi - x) f(\sin(\pi - x)) \, dx \]
Since \( \sin(\pi - x) = \sin x \), we can rewrite the integral as:
\[ I = \int_{0}^{\pi} (\pi - x) f(\sin x) \, dx \]

### Step 3: Simplify the integral
Now we can express \( I \) in terms of itself:
\[ I = \int_{0}^{\pi} \pi f(\sin x) \, dx - \int_{0}^{\pi} x f(\sin x) \, dx \]
This gives us:
\[ I = \pi \int_{0}^{\pi} f(\sin x) \, dx - I \]

### Step 4: Solve for \( I \)
Adding \( I \) to both sides, we get:
\[ 2I = \pi \int_{0}^{\pi} f(\sin x) \, dx \]
Thus,
\[ I = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \, dx \]

### Step 5: Final expression
The final expression for the integral is:
\[ I = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \, dx \]