Question
$\int_0^{2\pi } {\frac{{\sin 2\theta }}{{a - b\,\cos \theta }}\,d\theta = } $
✓
Answer
d
(d) $I = \int_0^{2\pi } {\frac{{\sin 2\theta }}{{a - b\cos \theta }}d\theta = \int_0^{2\pi } {\frac{{\sin (2\pi - 2\theta )}}{{a - b\cos (2\pi - \theta )}}d\theta } } $
(d) $I = \int_0^{2\pi } {\frac{{\sin 2\theta }}{{a - b\cos \theta }}d\theta = \int_0^{2\pi } {\frac{{\sin (2\pi - 2\theta )}}{{a - b\cos (2\pi - \theta )}}d\theta } } $
==> I $ = - \int_0^{2\pi } {\frac{{\sin 2\theta }}{{a - b\cos \theta }}d\theta } $
$ \Rightarrow \,\,2I = 0 $
$\Rightarrow \,\,\int_0^{2\pi } {\frac{{\sin 2\theta }}{{a - b\cos \theta }}d\theta = 0} $..
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