The value of \(\mathrm{I}=\int_{\sqrt{\log _e 2}}^{\sqrt{\log _e 3}} \frac{x \sin x^2}{\sin x^2+\sin \left(\log _e 6-x^2\right)} \mathrm{d} x\) is

Let \(\mathrm{I}=\int_{\sqrt{\log 2}}^{\sqrt{\log 3}} \frac{x \sin x^2}{\sin x^2+\sin \left(\log 6-x^2\right)} \mathrm{d} x\)
\(\begin{array}{ll}
& \text { Put } x^2=t \Rightarrow 2 x \mathrm{~d} x=\mathrm{dt} \\
\therefore \quad & \mathrm{I}=\frac{1}{2} \int_{\log _2}^{\log _3} \frac{\sin \mathrm{t}}{\sin \mathrm{t}+\sin (\log 6-\mathrm{t})} \mathrm{dt} ...(i)\\
\therefore \quad & \mathrm{I}=\frac{1}{2} \int_{\log 2}^{\log _2 3} \frac{\sin (\log 6-\mathrm{t})}{\sin (\log 6-\mathrm{t})+\sin \mathrm{t}} \mathrm{dt}...(ii)
\end{array}\)
\(\cdots\left[\because \int_a^b f(x) \mathrm{d} x=\int_a^b f(a+b-x) \mathrm{d} x\right]\)
Adding (i) and (ii), we get
\(\begin{aligned}
& 2 I=\frac{1}{2} \int_{\log 2}^{\log 3} \mathrm{dt}=\frac{1}{2}(\log 3-\log 2)=\frac{1}{2} \log \left(\frac{3}{2}\right) \\
& \Rightarrow I=\frac{1}{4} \log \left(\frac{3}{2}\right)
\end{aligned}\)