Помогите решить интеграл: [tex]\int\limits {sin^2xcos^6x} \, dx[/tex] Ответ: [tex]\frac{1}{16}(\frac{5}{8} x +\frac{1}{3} sin^32x-\frac{1}{8} sin4x - \frac{1}{64} sin8x)+C[/tex]
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