$\int$sin 4x sin 8x dx = $\frac{1}{2}$ $\int$2 sin 8x sin 4x dx $=\frac{1}{2}$ $\int$cos 4x - cos 12x)dx [$\because$ 2 sin A sin B = cos (A - B) - cos(A + B)] $=\frac{1}{2}\left(\frac{\sin 4 x}{4}-\frac{\sin 12 x}{12}\right)$ + C
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