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Integrate the...

Question

Integrate the following integrals:

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

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Solution

$\int \sin 2 x \sin 4 x \sin 6 x d x = \frac{1}{2} \int (2 \sin 2 x \sin 4 x) \sin 6 x d x = \frac{1}{2} \int [\cos (2 x - 4 x) - \cos (2 x + 4 x)] \sin 6 x d x = \frac{1}{2} \int [\cos (2 x) - \cos (6 x)] \sin 6 x d x = \frac{1}{2} [\int \cos (2 x) \sin (6 x) d x - \int \cos (6 x) \sin (6 x) d x] = \frac{1}{4} [\int 2 \cos (2 x) \sin (6 x) d x - \int 2 \cos (6 x) \sin (6 x) d x] = \frac{1}{4} \{\int [\sin (2 x + 6 x) - \sin (2 x - 6 x)] d x - \int \sin (12 x) d x\} = \frac{1}{4} [\int \sin (8 x) d x + \int \sin (4 x) d x - \int \sin (12 x) d x] = \frac{1}{4} [\frac{- \cos (8 x)}{8} + \frac{- \cos (4 x)}{4} + \frac{\cos (12 x)}{12}] + c = - \frac{\cos (8 x)}{32} - \frac{\cos (4 x)}{16} + \frac{\cos (12 x)}{48} + c$

Hence, $\int \sin 2 x \sin 4 x \sin 6 x d x = - \frac{\cos (8 x)}{32} - \frac{\cos (4 x)}{16} + \frac{\cos (12 x)}{48} + c$ .

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