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