求定积分x²cos2xdx上限为π下限为0

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∫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您好,土豆实力团为您答疑解难.
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======以下答案可供参考======
供参考答案1:
∫x²cos2xdx
=1/2∫x²cos2xd2x
=1/2∫x²dsin2x
=1/2x²sin2x-1/2∫sin2xdx²
=1/2x²sin2x+1/2∫xdcos2x
=1/2x²sin2x+1/2xcos2x-1/2∫cos2xdx
=1/2x²sin2x+1/2xcos2x-1/4sin2x (0,π
=-π供参考答案2:
求原函数啊，分布积分