Question
$\int_{}^{} {\frac{{\sin x\;dx}}{{{{(a + b\cos x)}^2}}} = } $
तब$\int_{}^{} {\frac{{\sin x}}{{{{(a + b\cos x)}^2}}}\,dx = - \frac{1}{b}\int_{}^{} {\frac{1}{{{t^2}}}\,dt} = \frac{1}{b}\frac{1}{t} + c} $
$ = \frac{1}{{b(a + b\cos x)}} + c.$
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