Find the integrals of the function cos 2x cos 4x cos 6x - Vidyadip

Question

Find the integrals of the function cos 2x cos 4x cos 6x

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Answer

Clearly,
$\int \cos 2 x \cos 4 x \cos 6 x d x$ = $\int \cos 2 x\left[\frac{1}{2}\{\cos (4 x+6 x)+\cos (4 x-6 x)\}\right] d x$
$=\frac{1}{2} \int\{\cos 2 x \cos 10 x+\cos 2 x \cos (-2 x)\} d x$
$= \frac{1}{2} \int\left\{\cos 2 x \cos 10 x+\cos ^{2} 2 x\right\} d x$
$= \frac{1}{2} \int\left[\frac{1}{2} \left\{\cos (2 \mathrm{x}+10 \mathrm{x})+\cos (2 \mathrm{x}-10 \mathrm{x})\right\}+\left(\frac{1+\cos 4 \mathrm{x}}{2}\right)\right] \mathrm{dx}$
$= \frac{1}{4} \int(\cos 12 x+\cos 8 x+1+\cos 4 x) d x$
$= \frac{1}{4}\left[\frac{\sin 12 \mathrm{x}}{12}+\frac{\sin 8 \mathrm{x}}{8}+\mathrm{x}+\frac{\sin 4 \mathrm{x}}{4}\right]+\mathrm{c}$

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