Definite Integration Question 268
Question: $ \int _{-\pi /2}^{\pi /2}{\log ( \frac{2-\sin \theta }{2+\sin \theta } )d\theta =} $
Options:
A) 0
B) 1
C) 2
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Since $ f(-\theta )=\log {{( \frac{2-\sin \theta }{2+\sin \theta } )}^{-1}}=-\log ( \frac{2-\sin \theta }{2+\sin \theta } )=-f(\theta ) $
$ \therefore $ $ f(x) $ is an odd function of $ x $ . Therefore, $ 2\int_0^{\pi /2}{\log ( \frac{2-\sin \theta }{2+\sin \theta } )d\theta =0} $ .
BETA
