If the integral `int_(0)^(2)(dx)/(sinx+sin(2-x))=A`, then the integral `beta=int_(0)^(2)(xdx)/(sinx+sin(2-x))` is equal to

Options

A. `(sin2)A`
B. 2A
C. A
D. `(A)/(2)`

Answer

To solve the integral \( \beta = \int_{0}^{2} \frac{x \, dx}{\sin x + \sin(2 - x)} \), we can use a property of definite integrals.

### Step 1: Set Up the Integral
We start with the integral:
\[ \beta = \int_{0}^{2} \frac{x \, dx}{\sin x + \sin(2 - x)} \]

### Step 2: Use the Property of Definite Integrals
We can use the property of integrals that states:
\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \]
In our case, \( a = 2 \) and we will evaluate \( f(2 - x) \):
\[ f(2 - x) = \frac{2 - x}{\sin(2 - x) + \sin x} \]
Thus, we can write:
\[ \beta = \int_{0}^{2} \frac{2 - x \, dx}{\sin(2 - x) + \sin x} \]

### Step 3: Combine the Integrals
Now we can combine the two expressions for \( \beta \):
\[ 2\beta = \int_{0}^{2} \left( \frac{x + (2 - x)}{\sin x + \sin(2 - x)} \right) dx \]
This simplifies to:
\[ 2\beta = \int_{0}^{2} \frac{2 \, dx}{\sin x + \sin(2 - x)} \]

### Step 4: Simplify the Integral
Now we can factor out the constant:
\[ 2\beta = 2 \int_{0}^{2} \frac{dx}{\sin x + \sin(2 - x)} \]
Dividing both sides by 2 gives us:
\[ \beta = \int_{0}^{2} \frac{dx}{\sin x + \sin(2 - x)} \]

### Step 5: Evaluate the Known Integral
From the problem statement, we know that:
\[ \int_{0}^{2} \frac{dx}{\sin x + \sin(2 - x)} = A \]
Thus, we have:
\[ \beta = A \]

### Final Answer
Therefore, the value of the integral \( \beta \) is equal to \( A \).