Question

The value of ${\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\log \left(\sec \left(\theta \right)-\tan \left(\theta \right)\right)dx$ is

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

Answer 1

The correct option is A

$0$

Explanation for the correct option:

Compute the required value:

Given: ${\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\log \left(\sec \left(\theta \right)-\tan \left(\theta \right)\right)dx$

Put $\theta =-\theta$

$$\begin{gathered} \Rightarrow \log \left(\sec \left(\theta \right)+\tan \left(\theta \right)\right) \\ \Rightarrow \log \frac{\left(\sec \left(\theta \right)+\tan \left(\theta \right)\right)}{1} \\ \Rightarrow \log \frac{\left(\sec \left(\theta \right)+\tan \left(\theta \right)\right)}{\left({\sec }^2\left(\theta \right)-{\tan }^2\left(\theta \right)\right)}\left[{\sec }^2\left(\theta \right)=1+{\tan }^2\left(\theta \right)\right] \\ \Rightarrow \log \frac{\left(\sec \left(\theta \right)+\tan \left(\theta \right)\right)}{\left(\sec \left(\theta \right)-\tan \left(\theta \right)\right)(\left(\sec \left(\theta \right)+\tan \left(\theta \right)\right)} \\ \Rightarrow \log \frac{1}{(\left(\sec \left(\theta \right)-\tan \left(\theta \right)\right)} \\ \Rightarrow -\log (\left(\sec \left(\theta \right)-\tan \left(\theta \right)\right) \end{gathered}$$

So, $\log \left(\sec \left(\theta \right)+\tan \left(\theta \right)\right)$ is an odd function.

${\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\log \left(\sec \left(\theta \right)-\tan \left(\theta \right)\right)dx$ is equal to $0$.

Hence, option $A$ is the correct answer.