int frac{sin 4x}{sin x} , dx = (where C is a | vedclass

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Question

(C) We have the integral $I = \int \frac{\sin 4x}{\sin x} \, dx$. Using the identity $\sin 4x = 2 \sin 2x \cos 2x = 4 \sin x \cos x \cos 2x$,we get: $

Vedclass · Accepted Answer

$\frac{2 \sin 3x}{3} + 2 \sin x + C$

Vedclass · Answer

(C) We have the integral $I = \int \frac{\sin 4x}{\sin x} \, dx$. Using the identity $\sin 4x = 2 \sin 2x \cos 2x = 4 \sin x \cos x \cos 2x$,we get: $I = \int \frac{4 \sin x \cos x \cos 2x}{\sin x} \, dx = 4 \int \cos x \cos 2x \, dx$. Using the formula $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,we have $2 \cos 2x \cos x = \cos(2x+x) + \cos(2x-x) = \cos 3x + \cos x$. Thus,$I = 2 \int (\cos 3x + \cos x) \, dx$. Integrating term by term,we get: $I = 2 \left( \frac{\sin 3x}{3} + \sin x \right) + C = \frac{2}{3} \sin 3x + 2 \sin x + C$.

$\int \frac{\sin 4x}{\sin x} \, dx =$ (where $C$ is a constant of integration.)

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