If the solve the problem

Question:

$\int \sin 2 x \sin 4 x \sin 6 x d x$

Solution:

We need to simplify the given equation to make it easier to solve

We know $2 \sin A \sin B=\cos (A-B)-\cos (A+B)$

$\therefore \sin 4 x \sin 2 x=\frac{\cos 2 x-\cos 6 x}{2}$

$\therefore$ The above equation becomes

$\Rightarrow \int \frac{1}{2}(\cos 2 x-\cos 6 x) \sin 6 x d x$

$\Rightarrow \frac{1}{2} \int((\cos 2 x \sin 6 x)-(\cos 6 x \sin 6 x)) d x$

We know $2 \sin A \cos B=\sin (A+B)+\sin (A-B)$

$\therefore \sin 6 x \cos 2 x=\frac{\sin 8 x+\sin 4 x}{2}$

Also $2 \sin x \cdot \cos x=\sin 2 x$

$\therefore \sin 6 x \cos 6 x=\frac{\sin 12 x}{2}$

$\therefore$ The above equation simplifies to

$\Rightarrow \frac{1}{2} \int \frac{1}{2}(\sin 8 x+\sin 4 x) d x-\int \frac{1}{2} \sin 12 x d x$

$\Rightarrow \frac{1}{4}\left(\int \sin 8 x d x+\int \sin 4 x d x-\int \sin 12 x d x\right)$

We know $\int \sin \mathrm{ax} \mathrm{dx}=\frac{-1}{\mathrm{a}} \cos \mathrm{ax}+c$

$\Rightarrow \frac{1}{4}\left(\frac{-1}{8} \cos 8 x+\frac{(-1)}{4} \cos 4 x+\frac{1}{12} \cos 12 x+c\right)$

$\Rightarrow \frac{1}{4}\left(\frac{2 \cos 12 x-3 \cos 8 x-6 \cos 4 x}{24}+c\right)$

$\Rightarrow \frac{2 \cos 12 x-3 \cos 8 x-6 \cos 4 x}{96}+c$ (where $c$ is some arbitrary constant)

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