• # question_answer 89) $\int_{\log \sqrt{\pi /2}}^{\log \sqrt{\pi }}{{{e}^{2x}}{{\sec }^{2}}\left( \frac{1}{3}{{e}^{2x}} \right)}dx$is equal to: A) $\sqrt{3}$                                        B) $\frac{1}{\sqrt{3}}$C) $\frac{3\sqrt{3}}{2}$                                   D) $\frac{1}{2\sqrt{3}}$

$I=\int_{\log \sqrt{\pi /2}}^{\log \sqrt{\pi }}{{{e}^{2x}}\sec }\left( \frac{1}{3}{{e}^{2x}} \right)dx$ put${{e}^{2x}}=t\Rightarrow 2{{e}^{2x}}dx=dt$ $x=\log \sqrt{\pi },\,\,t={{e}^{2\log \sqrt{\pi }}}=\pi$ When$x=\log \sqrt{\pi /2},\,\,t={{e}^{2\log \sqrt{\pi /2}}}={{e}^{\log \pi /2}}=\frac{\pi }{2}$ When$x=\log \sqrt{\pi },\,\,t={{e}^{2\log \sqrt{\pi }}}=\pi$ $\therefore$  $I=\int_{\frac{\pi }{2}}^{\pi }{{{\sec }^{2}}\left( \frac{1}{3}t \right)dt}=\frac{1}{2}\cdot \frac{1}{\frac{1}{3}}\left[ \tan \frac{t}{3} \right]_{\pi /2}^{\pi }$ $=\frac{3}{2}\left[ \tan \frac{\pi }{3}-\tan \frac{\pi }{6} \right]=\frac{3}{2}\left[ \sqrt{3}-\frac{1}{\sqrt{3}} \right]=\sqrt{3}$