int frac{sin 3x}{sin x} , dx = | vedclass

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(A) We know that $\sin 3x = 3\sin x - 4\sin^3 x$.Substituting this into the integral:$\int \frac{3\sin x - 4\sin^3 x}{\sin x} \, dx = \int (3 - 4\sin^

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$x + \sin 2x + c$

Vedclass · Answer

(A) We know that $\sin 3x = 3\sin x - 4\sin^3 x$.Substituting this into the integral:$\int \frac{3\sin x - 4\sin^3 x}{\sin x} \, dx = \int (3 - 4\sin^2 x) \, dx$Using the identity $\sin^2 x = \frac{1 - \cos 2x}{2}$:$\int (3 - 4(\frac{1 - \cos 2x}{2})) \, dx = \int (3 - 2(1 - \cos 2x)) \, dx$$= \int (3 - 2 + 2\cos 2x) \, dx = \int (1 + 2\cos 2x) \, dx$$= x + 2(\frac{\sin 2x}{2}) + c = x + \sin 2x + c$.

$\int \frac{\sin 3x}{\sin x} \, dx = $

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