$\[\frac{1}{{2\pi }}\mathop \smallint \limits_0^{2\pi } \frac{{{\rm{d}}\theta }}{{a + b\sin \theta }} = \frac{1}{{\sqrt {{a^2} - {b^2}} }}\]$
$\[\frac{1}{{2\pi }}\mathop \smallint \limits_0^{2\pi } \frac{{{\rm{d}}\theta }}{{a + b\sin \theta }} = \frac{1}{{\sqrt {{a^2} - {b^2}} }}\]$
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