Search any question & find its solution
Question:
Answered & Verified by Expert
$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin \theta}{2+\sin \theta}\right) d \theta$ is equal to
Options:
Solution:
2218 Upvotes
Verified Answer
The correct answer is:
$0$
Let $I=\int_{-\pi / 2}^{\pi / 2} \log \left(\frac{2-\sin \theta}{2+\sin \theta}\right) d \theta$
Let $\quad f(\theta)=\log \left(\frac{2-\sin \theta}{2+\sin \theta}\right)$
$\Rightarrow \quad f(-\theta)=\log \left(\frac{2+\sin \theta}{2-\sin \theta}\right)=-\log \left(\frac{2-\sin \theta}{2+\sin \theta}\right)$
$\Rightarrow \quad f(-\theta)=-f(\theta)$
$\therefore$ It is an odd function
$\therefore \quad I=\int_{-\pi / 2}^{\pi / 2} \log \left(\frac{2-\sin \theta}{2+\sin \theta}\right) d \theta=0$.
Let $\quad f(\theta)=\log \left(\frac{2-\sin \theta}{2+\sin \theta}\right)$
$\Rightarrow \quad f(-\theta)=\log \left(\frac{2+\sin \theta}{2-\sin \theta}\right)=-\log \left(\frac{2-\sin \theta}{2+\sin \theta}\right)$
$\Rightarrow \quad f(-\theta)=-f(\theta)$
$\therefore$ It is an odd function
$\therefore \quad I=\int_{-\pi / 2}^{\pi / 2} \log \left(\frac{2-\sin \theta}{2+\sin \theta}\right) d \theta=0$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.