Find the integrals of the function sin x sin 2x sin 3x - Vidyadip

Question

Find the integrals of the function sin x sin 2x sin 3x

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Answer

$\int \sin x \sin 2 x \sin 3 x d x$ = $\int \sin x \cdot \frac{1}{2}[\{\cos (2 x-3 x)-\cos (2 x+3 x)\}] d x$
$= \frac{1}{2} \int\{\sin x \cos (-x)-\sin x \cos 5 x\} d x$
$= \frac{1}{2} \int\{\sin x \cos x-\sin x \cos 5 x\} d x$
$= \frac{1}{2} \int \frac{\sin 2 x}{2} d x-\frac{1}{2} \int \sin x \cos 5 x d x$
$= \frac{1}{4}\left[\frac{-\cos 2 x}{2}\right]-\frac{1}{2} \int \frac{1}{2}\left\{ \sin (x+5 x)+\sin (x-5 x)\right\} d x$
$= \frac{-\cos 2 x}{8}-\frac{1}{4} \int(\sin 6 x+\sin (-4 x)) d x$
$= \frac{-\cos 2 x}{8}-\frac{1}{4}\left[\frac{-\cos 6 x}{6}+\frac{\cos 4 x}{4}\right]+C$
$= \frac{-\cos 2 x}{8}-\frac{1}{8}\left[\frac{-\cos 6 x}{3}+\frac{\cos 4 x}{2}\right]+C$
$= \frac{1}{8}\left[\frac{\cos 6 x}{3}-\frac{\cos 4 x}{2}-\cos 2 x\right]+C$

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