求定积分x²cos2xdx上限为π下限为0

∫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!

∫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

求原函数啊，分布积分