∫x^2 cosx dx 0~2π 定积分求解~~~~

2025-06-21 16:53:58
推荐回答(1个)
回答1:

∫x^2 cosx dx
=∫x²dsinx
=x²sinx-∫sinxdx²
=x²sinx-2∫xsinxdx
=x²sinx+2∫xdcosx
=x²sinx+2xcosx-2∫cosxdx
=x²sinx+2xcosx-2sinx (0~2π)
=(0+4π-0)-(0+0+0)
=4π