A) 0
B) 1
C) 2
D) None of these
Correct Answer: A
Solution :
Since \[f(-\theta )=\log {{\left( \frac{2-\sin \theta }{2+\sin \theta } \right)}^{-1}}=-\log \left( \frac{2-\sin \theta }{2+\sin \theta } \right)=-f(\theta )\] \[\therefore \] \[f(x)\] is an odd function of \[x\]. Therefore, \[2\int_{0}^{\pi /2}{\log \left( \frac{2-\sin \theta }{2+\sin \theta } \right)\text{ }d\theta =0}\].You need to login to perform this action.
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