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

Put \(x^2=t \Rightarrow x d x=\frac{d t}{2}\) \(\therefore \quad I=\int_{\log 2}^{\log 3} \frac{\sin t \cdot \frac{d t}{2}}{\sin t+\sin (\log 6-t)}\)
Using, \(\quad \int_a^b f(x) d x=\int_a^b f(a+b-x) d x\)
\[
\begin{aligned}
& I=\frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin (\log 2+\log 3-t)}{\sin (\log 2+\log 3-t)+\sin } d t \\
& I=\frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin (\log 6-t)}{\sin (\log 6-t)+\sin (t)} d t \\
& \therefore I=\int_{\log 2}^{\log 3} \frac{\sin (\log 6-t)}{\sin (\log 6-t)+\sin t} d t \quad \ldots(\mathrm{li})
\end{aligned}
\]
On adding Eqs. (i) and (ii), we get
\[
2 I=\frac{1}{2} \int_{\log 2}^{\log 3} \frac{\sin t+\sin (\log 6-t)}{\sin (\log 6-t)+\sin t} d t
\]
\[
\begin{gathered}
\therefore \quad 2 I=\frac{1}{2}(t)_{\log 2}^{\log 3}=\frac{1}{2}(\log 3-\log 2) \\
\Rightarrow I=\frac{1}{4} \log \left(\frac{3}{2}\right)
\end{gathered}
\]