求定积分x²cos2xdx上限为π下限为0

3个回答

  • ∫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

    您好,土豆实力团为您答疑解难.

    如果本题有什么不明白可以追问,如果满意记得采纳.

    答题不易,请谅解,谢谢.

    另祝您学习进步!