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求定积分x²cos2xdx上限为π下限为0
人气:433 ℃ 时间:2020-02-06 12:26:06
解答
∫x²cos2xdx
=1/2·∫x²dsin2x
=1/2·x²sin2x-1/2·∫sin2xdx²
=1/2·x²sin2x-∫xsin2xdx
=1/2·x²sin2x+1/2∫xdcos2x
=1/2·x²sin2x+1/2xcos2x-1/2∫cos2xdx
=1/2·x²sin2x+1/2xcos2x-1/4∫dsin2x
=1/2·x²sin2x+1/2xcos2x-1/2sin2x
所以求定积分x²cos2xdx上限为π下限为0
=(1/2·x²sin2x+1/2xcos2x-1/2cos2x) |(0到π)
=-π

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