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Evaluate the following indefinite integral : ∫1/(sin x - sin 2x) dx.

\(\int\cfrac{1}{sin\,x-sin\,2x}dx.\)

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∫1/(sin x - sin 2x) dx

 =\(\int\cfrac{1}{sin\,x-sin\,2x}dx\) = \(\int\cfrac{1}{sin\,x-2sin\,x\,cos\,x}dx\) (\(\because\) sin 2x = 2sin x cosx)

(\(\because\) sin2x = 2 sin x cos x)

Where, C is an integral constant. Now, putting t = cos x, we get

Where C is an integral constant.

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