The valued of `int_(sqrt(In2))^(sqrt(In3)) (x sinx^(2))/(sinx^(2)+sin(In 6-x^(2)))dx` is

Options

A. `1/4 In 3/2`
B. `1/2 In 3/2`
C. `In 3/2`
D. `1/6 In 3/2`

Explanation

To solve the integral

\[ I = \int_{\sqrt{\ln 2}}^{\sqrt{\ln 3}} \frac{x \sin(x^2)}{\sin(x^2) + \sin(\ln 6 - x^2)} \, dx, \]

we will follow these steps:

### Step 1: Substitution
Let \( x^2 = t \). Then, differentiating both sides gives \( 2x \, dx = dt \) or \( dx = \frac{dt}{2\sqrt{t}} \).

### Step 2: Change of Limits
When \( x = \sqrt{\ln 2} \), \( t = \ln 2 \) and when \( x = \sqrt{\ln 3} \), \( t = \ln 3 \).

### Step 3: Rewrite the Integral
Now substituting \( x^2 = t \) into the integral, we have:

\[ I = \int_{\ln 2}^{\ln 3} \frac{\sqrt{t} \sin t}{\sin t + \sin(\ln 6 - t)} \cdot \frac{dt}{2\sqrt{t}} = \frac{1}{2} \int_{\ln 2}^{\ln 3} \frac{\sin t}{\sin t + \sin(\ln 6 - t)} \, dt. \]

### Step 4: Use the Property of Definite Integrals
Using the property of definite integrals, we have:

\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx. \]

In our case, let \( a = \ln 2 \) and \( b = \ln 3 \). Thus:

\[ I = \frac{1}{2} \int_{\ln 2}^{\ln 3} \frac{\sin t}{\sin t + \sin(\ln 6 - t)} \, dt = \frac{1}{2} \int_{\ln 2}^{\ln 3} \frac{\sin(\ln 6 - t)}{\sin(\ln 6 - t) + \sin t} \, dt. \]

### Step 5: Adding the Two Integrals
Now, adding the two integrals we have:

\[ 2I = \frac{1}{2} \int_{\ln 2}^{\ln 3} \left( \frac{\sin t}{\sin t + \sin(\ln 6 - t)} + \frac{\sin(\ln 6 - t)}{\sin(\ln 6 - t) + \sin t} \right) dt. \]

This simplifies to:

\[ 2I = \frac{1}{2} \int_{\ln 2}^{\ln 3} 1 \, dt = \frac{1}{2} \cdot \left( \ln 3 - \ln 2 \right). \]

### Step 6: Solve for I
Thus, we have:

\[ 2I = \frac{1}{2} (\ln 3 - \ln 2) \implies I = \frac{1}{4} (\ln 3 - \ln 2) = \frac{1}{4} \ln \left( \frac{3}{2} \right). \]

### Final Answer
Therefore, the value of the integral is

\[ I = \frac{1}{4} \ln \left( \frac{3}{2} \right). \]