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

$\int \frac{\sin 4 x}{\sin x} \mathrm{~d} x=$ (where $C$ is a constant of integration.)
  1. $\frac{\sin 3 x}{3}+4 \sin x+C$
  2. $\frac{1}{3} \sin 3 x-\frac{2}{3} \sin x+C$
  3. $\frac{2 \sin 3 x}{3}+2 \sin x+C$
  4. $\frac{2}{3} \sin 3 x-2 \sin x+C$

Solution

$\begin{aligned} & \int \frac{\sin 4 x}{\sin x} \mathrm{~d} x=\int \frac{4 \sin x \cdot \cos x \cdot \cos 2 x}{\sin x} \mathrm{~d} x=2 \int 2 \cos x \cdot \cos 2 x d x \\ & =2 \int\{\cos 3 x+\cos x\} \mathrm{d} x \\ & =2\left\{\frac{\sin 3 x}{3}+\sin x\right\}+c \\ & =\frac{2}{3} \sin 3 x+2 \sin x+c\end{aligned}$

Asked in: MHT CET 2022 (08 Aug Shift 1)

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