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求助

x3sinx2的不定积分

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IP属地:山东1楼2023-05-12 15:22回复
    解∫x³sinx²dx=1/2∫x²sinx²dx²=1/2∫usinudu=-1/2∫ud(cosu)=-1/2[ucosu-∫cosudu]=-1/2ucosu+1/2sinu+C=-1/2x²cosx²+1/2sinx²+C


    IP属地:上海2楼2023-05-26 01:03
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      解∫x³sinx²dx=1/2∫x²sinx²dx²=1/2∫usinudu=-1/2∫ud(cosu)=-1/2[ucosu-∫cosudu]=-1/2ucosu+1/2sinu+C=-1/2x²cosx²+1/2sinx²+C


      IP属地:安徽3楼2023-05-26 01:03
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