∫x눀sin대xdx换元怎么解

2024-12-02 10:11:27
推荐回答(2个)
回答1:

∫x²sin³xdx=1/4∫x²(3sinx-sin3x)dx
=3/4∫x²sinxdx-1/4∫x²sin3xdx
=-3/4∫x²dcosx+1/12∫x²dcos3x
=-3/4x²cosx+3/4∫cosxdx²+1/12x²cos3x-1/12∫cos3xdx²
=-3/4x²cosx+3/2∫xcosxdx+1/12x²cos3x-1/6∫xcos3xdx
=-3/4x²cosx+3/2∫xdsinx+1/12x²cos3x-1/18∫xdsin3x
=-3/4x²cosx+3/2xsinx-3/2∫sinxdx +1/12x²cos3x-1/18xsin3x+1/18∫sin3xdx
=-3/4x²cosx+3/2xsinx+3/2cosx +1/12x²cos3x-1/18xsin3x-1/54cosx+C

∫x²sinx³dx=1/3∫sinx³dx³=-1/3cosx³+C

回答2:

sin 三次方错位了吧