int_0^{2pi } {frac{{sin 2theta }}{{a - bcos t | vedclass

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(D) Let $I = \int_0^{2\pi } {\frac{{\sin 2	heta }}{{a - b\cos 	heta }}d	heta } $. Using the property $\int_0^{2a} f(x) dx = \int_0^{2a} f(2a-x) dx$

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(D) Let $I = \int_0^{2\pi } {\frac{{\sin 2	heta }}{{a - b\cos 	heta }}d	heta } $. Using the property $\int_0^{2a} f(x) dx = \int_0^{2a} f(2a-x) dx$,we have: $I = \int_0^{2\pi } {\frac{{\sin (2(2\pi - 	heta ))}}{{a - b\cos (2\pi - 	heta )}}d	heta } $ $I = \int_0^{2\pi } {\frac{{\sin (4\pi - 2	heta )}}{{a - b\cos 	heta }}d	heta } $ Since $\sin(4\pi - 2	heta) = -\sin 2	heta$,we get: $I = - \int_0^{2\pi } {\frac{{\sin 2	heta }}{{a - b\cos 	heta }}d	heta } $ $I = -I$ $2I = 0$ $I = 0$.

$\int_0^{2\pi } {\frac{{\sin 2\theta }}{{a - b\cos \theta }}\,d\theta = } $

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