Question

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

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

Answer 1

Answer: (A)

$0$

$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\log \left(\frac{2-sin\theta }{2+sin\theta }\right)d\theta$ ...(1)
Using property ${\int }_{-a}^{a}f\left(x\right)dx=0$, if $f\left(x\right)$ is odd function.
Now let $g\left(x\right)=log\left(\frac{2-sin\theta }{2+sin\theta }\right)$, which is an odd function.
Therefore,
$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\log \left(\frac{2-sin\theta }{2+sin\theta }\right)d\theta =0$
Hence, option 'A' is correct.