Question

The value of the integral $\underset{-\pi /4}{\overset{\pi /4}{\int }}\log (\sec \theta -\tan \theta )d\theta$ is

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $0$
• D. $\pi$

Answer 1

Answer: (C)

Let $I=\underset{-\pi /4}{\overset{\pi /4}{\int }}\log (\sec \theta -\tan \theta )d\theta$
Again, let $f(\theta )=\log (\sec \theta -\tan \theta )$
$\therefore f(-\theta )=\log [\sec (-\theta )-\tan (-\theta )]$
$=\log \left[(\sec \theta +\tan \theta )\times \frac{\sec \theta -\tan \theta }{\sec \theta -\tan \theta }\right]$
$=\log \left[\frac{{\sec }^2\theta -{\tan }^2\theta }{\sec \theta -\tan \theta }\right]=\log \left[\frac{1}{\sec \theta -\tan \theta }\right]$
$=\log 1-\log (\sec \theta -\tan \theta )$
$=0-\log (\sec \theta -\tan \theta )$
$\Rightarrow f(-\theta )=-f(\theta )$
Hence, $f(\theta )$ is an odd function.
$\therefore I=0$