A) 0
B) 1
C) 2
D) None of these
Correct Answer: A
Solution :
Let \[I=\int_{-\pi /2}^{\pi /2}{\log \left( \frac{2-\sin \theta }{2+\sin \theta } \right)}d\theta \] put \[\theta =-\theta \] \[I=\int_{-\pi /2}^{\pi /2}{\log \left( \frac{2-\sin (-\theta )}{2+\sin (-\theta )} \right)}\,d\theta \] \[=\int_{-\pi /2}^{\pi /2}{\log \left( \frac{2+\sin \theta }{2-\sin \theta } \right)}\,d\theta \] \[=\int_{-\pi /2}^{\pi /2}{-\log \left( \frac{2-\sin \theta }{2+\sin \theta } \right)}\,d\theta \] \[=-I\] \[\therefore \] \[I=0\]
You need to login to perform this action.
You will be redirected in
3 sec
