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为什么sin2xCos3x=(sin5x%sinx)/2,各位仁兄能说说...

根据公式sinacosb=1/2(sin(a+b)+sin(a-b))得 sin2xcos3x=1/2(sin(2x+3x)+sin(2x-3x)) =1/2(sin5x+sin(-x)) =(sin5x-sinx)/2

sin2xcos3x =1/2[sin(2x+3x)+sin(2x-3x)] =1/2(sin5x-sinx)

积化和差公式: sinαcosβ=1/2[sin(α+β)+sin(α-β)] 本题中,将3x看做α,2x看做β: sin3xcos2x = 1/2[sin(3x+2x)+sin(3x-2x)] = 1/2(sin5x+sinx)

先积化和差 cos2xcos3x = (1/2)[cos5x + cosx] 再积分得 (1/10)sin5x + (1/2)sinx + C

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