∫x/sin^3dx是否可积 20
1个回答
展开全部
∫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
= ∫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
追问
是x除以sin^3
本回答被网友采纳
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询