$\mathrm{I}=\int_{\log \sqrt{\pi / 2}}^{\log \sqrt{\pi}} \mathrm{e}^{2 \mathrm{x}} \sec ^2\left(\frac{1}{3} \mathrm{e}^{2 \mathrm{x}}\right) \mathrm{dx}$
Put $\mathrm{e}^{2 \mathrm{x}}=\mathrm{t} \Rightarrow 2 \mathrm{e}^{2 \mathrm{x}} \mathrm{dx}=\mathrm{dt}$
When
$$
\begin{aligned}
& x=\log \sqrt{\pi / 2}, t=e^{2 \log \sqrt{\pi / 2}} \\
& =\mathrm{e}^{\log \pi / 2}=\frac{\pi}{2} \\
& \text { When } \mathrm{x}=\log \sqrt{\pi}, \mathrm{t}=\mathrm{e}^{2 \log \sqrt{\pi}}=\pi \\
&
\end{aligned}
$$


$$
\begin{aligned}
\therefore \mathrm{I} & =\int_{\frac{\pi}{2}}^\pi \frac{1}{2} \sec ^2\left(\frac{1}{3} \mathrm{t}\right) \mathrm{dt}=\frac{1}{2} \cdot \frac{1}{\frac{1}{3}}\left[\tan \frac{\mathrm{t}}{3} \cdot\right]_{\pi / 2}^\pi \\
& =\frac{3}{2}\left[\tan \frac{\pi}{3}-\tan \frac{\pi}{6}\right]=\frac{3}{2}\left[\sqrt{3}-\frac{1}{\sqrt{3}}\right]=\sqrt{3}
\end{aligned}
$$