Evaluate Definite Integral Involving Sine and Logarithmic Terms

To evaluate the given definite integral:

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

Step 1: Substitution
Let $t = x^2$. Differentiating both sides gives:
$dt = 2x \, dx \implies x \, dx = \frac{dt}{2}$

Now, change the limits of integration:

• Lower limit: When $x = \sqrt{\ln 2}$, $t = (\sqrt{\ln 2})^2 = \ln 2$.
• Upper limit: When $x = \sqrt{\ln 3}$, $t = (\sqrt{\ln 3})^2 = \ln 3$.

Substituting these into the integral gives:
$I = \frac{1}{2} \int_{\ln 2}^{\ln 3} \frac{\sin t}{\sin t + \sin (\ln 6 - t)} \, dt \quad \text{--- (1)}$

Step 2: Apply the property of definite integrals
Using King's Property of definite integrals, $\int_{a}^{b} f(t) \, dt = \int_{a}^{b} f(a + b - t) \, dt$:
Here, the sum of the limits is $a + b = \ln 2 + \ln 3 = \ln(2 \cdot 3) = \ln 6$.

Replacing $t$ with $(\ln 6 - t)$ in equation (1):
$I = \frac{1}{2} \int_{\ln 2}^{\ln 3} \frac{\sin (\ln 6 - t)}{\sin (\ln 6 - t) + \sin (\ln 6 - (\ln 6 - t))} \, dt$

$I = \frac{1}{2} \int_{\ln 2}^{\ln 3} \frac{\sin (\ln 6 - t)}{\sin (\ln 6 - t) + \sin t} \, dt \quad \text{--- (2)}$

Step 3: Add equation (1) and equation (2)
$2I = \frac{1}{2} \int_{\ln 2}^{\ln 3} \left( \frac{\sin t + \sin (\ln 6 - t)}{\sin t + \sin (\ln 6 - t)} \right) \, dt$

$2I = \frac{1}{2} \int_{\ln 2}^{\ln 3} 1 \, dt$

$2I = \frac{1}{2} \left[ t \right]_{\ln 2}^{\ln 3}$

$2I = \frac{1}{2} (\ln 3 - \ln 2)$

$2I = \frac{1}{2} \ln \left(\frac{3}{2}\right)$

$I = \frac{1}{4} \ln \left(\frac{3}{2}\right)$

Thus, the correct option is (A).