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Integrate the following w. r. t. x :

\(\frac{1}{sin\,x + sin\,2x}\)

1/(sin x + sin 2x)

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Best answer

I = 1/(sin x + sin 2x)dx

Put cos x = t

∴ -sin x dx = dt

∴ sin x dx = -dt

Putting 1 - t = 0, i.e. t = 1, we get

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