The value of $\int_{\sqrt{\ln 2}}^{\sqrt{\ln 3}} \frac{x \sin x^{2}}{\sin x^{2} + \sin(\ln 6 - x^{2})}\,dx$ is

The value of ∫_(√ln 2)^(√ln 3) (x sin x²)/(sin x² + sin(ln 6 - x²)) dx is

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

$\text{Evaluate } I$

$1.\ \text{substitute } t = x^{2}$

$2.\ \text{find the symmetry of the integrand}$

$3.\ \text{add two equal integrals}$

$\left(\sqrt{\ln2}\right)^2 = \ln2, \quad \left(\sqrt{\ln3}\right)^2 = \ln3$

$\ln2 + \ln3 = \ln6$

$t = x^{2}, \quad dt = 2x\,dx \Rightarrow x\,dx = \dfrac{dt}{2}$

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

$g(t) = \dfrac{\sin t}{\sin t + \sin(C - t)}, \quad C = \ln 6$

$g(C - t) = 1 - g(t) \ \Rightarrow\ g(t) + g(C-t) = 1$

$\displaystyle\int_{\ln2}^{\ln3} g(t)\,dt = \displaystyle\int_{\ln2}^{\ln3} g(C-t)\,dt$

$2\displaystyle\int_{\ln2}^{\ln3} g(t)\,dt = \displaystyle\int_{\ln2}^{\ln3}\big[g(t)+g(C-t)\big]dt = \displaystyle\int_{\ln2}^{\ln3} 1\,dt$

$= \ln3 - \ln2 = \ln\dfrac{3}{2}$

$\Rightarrow \displaystyle\int_{\ln2}^{\ln3} g(t)\,dt = \dfrac{1}{2}\ln\dfrac{3}{2}$

$I = \dfrac{1}{2}\cdot\dfrac{1}{2}\ln\dfrac{3}{2} = \dfrac{1}{4}\ln\dfrac{3}{2}$

$\Rightarrow \text{option (1)}$