Question

The value of ${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\log \left(\frac{2-\sin \theta }{2+\sin \theta }\right)d\theta$ is

• A. $0$
• B. $1$
• C. $2$
• D. none of these

Answer 1

The correct option is A

$0$

Explanation for the correct option:

Compute the required value:

Given: ${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\log \left(\frac{2-\sin \theta }{2+\sin \theta }\right)dx$

$\Rightarrow f\left(\theta \right)=\log \frac{2-\sin \theta }{2+\sin \theta }$

Put $\theta =-\theta$

$\begin{array}{rcl}f(-\theta )& =& \log \left(\frac{2-\sin (-\theta )}{2+\sin (-\theta )}\right)\\ & =& \log \left(\frac{2+\sin \left(\theta \right)}{2-\sin \left(\theta \right)}\right)\\ & =& \log {\left(\frac{2-\sin \left(\theta \right)}{2+\sin \left(\theta \right)}\right)}^{-1}\\ & =& -\log \left(\frac{2-\sin \left(\theta \right)}{2+\sin \left(\theta \right)}\right)\\ f(-\theta )& =& -f\left(\theta \right)\end{array}$

So, $f\left(\theta \right)=\log \left(\frac{2-\sin \left(\theta \right)}{2+\sin \left(\theta \right)}\right)$ is an odd function

${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\log \left(\frac{2-\sin \theta }{2+\sin \theta }\right)d\theta$ is equal to $0$.

Hence, option $A$ is the correct answer.