The value of `int _(sqrt ( log 2 )) ^(sqrt( log 3 )) (x sin x ^(2))/( sin x ^(2) + sin (log 6 -x ^(2)))dx` is

Options

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

Answer

To solve the integral

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

we will follow the steps outlined in the video transcript.

### Step 1: Change of Variable
Let \( t = x^2 \). Then, \( dx = \frac{1}{2\sqrt{t}} \, dt \). The limits change as follows:
- When \( x = \sqrt{\log 2} \), \( t = \log 2 \).
- When \( x = \sqrt{\log 3} \), \( t = \log 3 \).

Thus, the integral becomes:

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

### Step 2: Use Symmetry
Using the property of definite integrals, we can express \( I \) as:

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

Now, let's consider the integral:

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

By substituting \( u = \log 6 - t \), we find that the limits change accordingly, and we can show that this integral is equal to \( I \).

### Step 3: Combine Integrals
Now we have:

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

This simplifies to:

\[ 2I = \int_{\log 2}^{\log 3} dt = \log 3 - \log 2 = \log\left(\frac{3}{2}\right). \]

### Step 4: Solve for \( I \)
Thus, we have:

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

### Final Answer
The value of the integral is:

\[ \boxed{\frac{1}{2} \log\left(\frac{3}{2}\right)}. \]

---