Question

Evaluate: $\int\limits^{\pi/2}_{-\pi/2} {\log (\frac{2-\sin x}{2+\sin x})} \, dx$

Answer 1

Solution:

let
$f(x) = log( \frac{2 - \sin(x) }{2 + \sin(x) } ) \\ \\ then \\ \\ f( - x) = log( \frac{2 + \sin(x) }{2 - \sin(x) } ) \\ \\ = log ({ \frac{2 - sin \: x}{2 - sin \: x} })^{ - 1} \\ \\ \\ = - log( \frac{2 - \sin(x) }{2 + \sin(x) } ) \\ \\ = - f(x)$
because f(x) is an odd function of x

and we know that

$\int\limits^{a}_{-a}f(x) \: dx = 0$
when f(x) is an odd function of x

so

$\int\limits^{\pi/2}_{-\pi/2} {\log (\frac{2-\sin x}{2+\sin x})} \, dx = 0$