Find the integrals of the functions.
$\int$ sin x sin 2x sin 3x dx

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Solution

$\int s i n x s i n 2 x s i n 3 x d x = \frac{1}{2} \int (2 s i n 3 x s i n x) s i n 2 x d x [∵ 2 s i n A s i n B = c o s (A - B) - c o s (A + B)] = \frac{1}{2} \int (c o s 2 x - c o s 4 x) s i n 2 x d x = \frac{1}{4} \int 2 s i n 2 x c o s 2 x - 2 c o s 4 x s i n 2 x d x = \frac{1}{4} \int (s i n 4 x - s i n 0) - (s i n 6 x - s i n 2 x) d x [\begin{matrix} ∵ 2 s i n A c o s B = s i n (A + B) + s i n (A - B) 2 c o s A s i n B = s i n (A + B) - s i n (A - B) \end{matrix}] \frac{1}{4} \int (s i n 4 x - s i n 6 x + s i n 2 x) d x = \frac{1}{4} {\frac{- c o s 4 x}{4} - \frac{(- c o s 6 x)}{6} + \frac{(- c o s 2 x)}{2}} + C$

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