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∫x²sin³xDx换元怎么解

∫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+1/12...

∫sin³xdx =∫sinx(1-cos²x)dx =∫(sinx-sinxcos²x)dx =-cosx+∫cos²xd(cosx) =-cosx-cos³x/3+C

∫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+1/12...

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