Evaluate: sin x sin 2x sin 3x dx. - Vidyadip

Question

Evaluate: $\int$sin x sin 2x sin 3x dx.

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Answer

$\text{I}=\int\text{sin x sin 2x sin 3x dx}=\frac{1}{2}\int\text{2sin 3x sin x sin 2x dx}$

$=\frac{1}{2}\int\text{(cos 2x - cos 4x) sin 2x dx}=\frac{1}{2}\int\text{(sin 2x cos 2x -cos 4x sin 2x)dx}$

$=\frac{1}{4}\int\text{sin 4x dx}-\frac{1}{4}\int\text{2 cos 4x sin 2x dx}$

$=-\frac{1}{16}\int\text{cos 4x }-\frac{1}{4}\int\text{(sin 6x - sin 2x)dx}$

$=-\frac{1}{16}\text{cos 4x}+\frac{1}{24}\text{cos 6x}-\frac{1}{8}\text{cos 2x + c}$

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