Question

∫x.(sinx)^3dx 求不定积分

Answer 1

分都没有，我靠
原式=-∫x(1-cosx^2)dcosx=-∫xdcosx+∫xcosx^2dcosx
后面一部分=1/3xcosx^3-1/3∫cosx^3dx
再往下就不用我说了吧，给分

Answer 2

∫xsin³x dx
= ∫x(1-cos²x)sinx dx
= ∫xsinx dx - ∫xcos²xsinx dx
= ∫xsinx dx - (1/2)∫x(1+cos2x)sinx dx
= ∫xsinx dx - (1/2)∫xsinx dx - (1/2)∫xsinxcos2x dx
= (1/2)∫xsinx dx - (1/4)∫x(sin3x-sinx) dx
= (1/2)∫xsinx dx - (1/4)∫xsin3x dx + (1/4)∫xsinx dx
= -(3/4)∫x dcosx + (1/4)(1/3)∫x dcos3x
= -(3/4)xcosx + (3/4)∫cosx dx + (1/12)xcos3x - (1/12)∫cos3x dx
= -(3/4)xcosx + (3/4)sinx + (1/12)xcos3x - (1/36)sin3x + C