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Решите задачу:

$\int \, sin^4x\, cos^4x\, dx=\int \, (sinx\cdot cosx)^4\, dx=\int (\frac{1}{2}sin2x)^4dx= \\\\=\frac{1}{16}\int (sin^22x)^2\, dx= \frac{1}{16}\int (\frac{1-cos4x}{2})^2\, dx=\frac{1}{64}\int (1-2cos4x+cos^24x)\, dx=\\\\=\frac{1}{64}\int (1-2cos4x+\frac{1+cos8x}{2})\, dx=\frac{1}{64}\int (\frac{3}{2}-2cos4x+\frac{1}{2}cos8x)\, dx=\\\\=\frac{1}{64}\, (\frac{3}{2}x-\frac{2}{4}sin4x+\frac{1}{16}sin8x)+C=\frac{1}{64}\, (\frac{3}{2}x-\frac{1}{2} sin4x+\frac{1}{16}sin8x)+C$