MCQ
$\int_{ - \pi /2}^{\pi /2} {\log \left( {\frac{{2 - \sin \theta }}{{2 + \sin \theta }}} \right)\,d\theta = } $
  • $0$
  • B
    $1$
  • C
    $2$
  • D
    None of these

Answer

Correct option: A.
$0$
a
(a) 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){\rm{ }}d\theta = 0} $.

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