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 problem, we need to evaluate the integral

\[ \beta = \int_{0}^{2} \frac{x \, dx}{\sin x + \sin(2 - x)} \]

given that

\[ A = \int_{0}^{2} \frac{dx}{\sin x + \sin(2 - x)}. \]

### Step 1: Understanding the Integral

We start with the integral \( A \):

\[ A = \int_{0}^{2} \frac{dx}{\sin x + \sin(2 - x)}. \]

### Step 2: Using Symmetry

Notice that \( \sin(2 - x) = \sin x \) due to the sine function's periodicity and symmetry. Thus, we can rewrite the integral:

\[ A = \int_{0}^{2} \frac{dx}{\sin x + \sin(2 - x)} = \int_{0}^{2} \frac{dx}{\sin x + \sin x} = \int_{0}^{2} \frac{dx}{2 \sin x}. \]

### Step 3: Setting Up the Integral for \(\beta\)

Now, we will evaluate \( \beta \):

\[ \beta = \int_{0}^{2} \frac{x \, dx}{\sin x + \sin(2 - x)}. \]

### Step 4: Change of Variable

We can use the substitution \( x = 2 - u \). Then, \( dx = -du \), and the limits change as follows:

- When \( x = 0 \), \( u = 2 \).
- When \( x = 2 \), \( u = 0 \).

Thus, we have:

\[ \beta = \int_{2}^{0} \frac{(2 - u)(-du)}{\sin(2 - u) + \sin u} = \int_{0}^{2} \frac{(2 - u) \, du}{\sin(2 - u) + \sin u}. \]

### Step 5: Simplifying the Integral

Now we can express the integral:

\[ \beta = \int_{0}^{2} \frac{(2 - u) \, du}{\sin(2 - u) + \sin u} = \int_{0}^{2} \frac{2 \, du}{\sin(2 - u) + \sin u} - \int_{0}^{2} \frac{u \, du}{\sin(2 - u) + \sin u}. \]

Notice that the second integral is just \( \beta \):

\[ \beta = 2 \int_{0}^{2} \frac{du}{\sin(2 - u) + \sin u} - \beta. \]

### Step 6: Solving for \(\beta\)

Adding \( \beta \) to both sides gives:

\[ 2\beta = 2A \implies \beta = A. \]

### Final Result

Thus, we conclude that:

\[ \beta = A. \]