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∫x²;·cos²;(x/2)dx=∫x²;·(cosx+1)dx=∫x²;cosxdx+∫x²;dx=∫x²;cosxdx+x³;/3①下麵求∫x²;cosxdx∫x²;cosxdx=x²;sinx-∫2xsinxdx=x²;sinx-[-2xcosx+∫2cosxdx]…
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