Question

${\int }_{\log \sqrt{\pi /2}}^{\log \sqrt{\pi }}{e}^{2x}{\sec }^2\left(\frac{1}{3}{e}^{2x}\right)dx$ is equal to

• A. $\sqrt{3}$
• B. $\frac{1}{\sqrt{3}}$
• C. $\frac{3\sqrt{3}}{2}$
• D. $\frac{1}{2\sqrt{3}}$

Answer 1

Answer: (A)

$<br/>I={\int }_{\log \sqrt{\pi /2}}^{\log \sqrt{\pi }}{e}^{2x}{\sec }^2\left(\frac{1}{3}{e}^{2x}\right)dx<br/>$
Put $e^{2x}=t\Rightarrow 2e^{2x}dx=dt$
When $x=\log \sqrt{\pi /2},t=e^{2\log \sqrt{\pi /2}}=e^{\log \pi /2}=\frac{\pi }{2}$
When $x=\log \sqrt{\pi },t=e^{2\log \sqrt{\pi }}=\pi$
$<br/>\begin{array}{l}<br/>\therefore I={\int }_{\frac{\pi }{2}}^{\pi }\frac{1}{2}{\sec }^2\left(\frac{1}{3}t\right)dt=\frac{1}{2}\cdot \frac{1}{\frac{1}{3}}{\left[\tan \frac{t}{3}\cdot \right]}_{\pi /2}^{\pi }\\ <br/>=\frac{3}{2}\left[\tan \frac{\pi }{3}-\tan \frac{\pi }{6}\right]=\frac{3}{2}\left[\sqrt{3}-\frac{1}{\sqrt{3}}\right]=\sqrt{3}<br/>\end{array}<br/>$