∫x/sin^3dx是否可积
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解决时间 2021-03-31 19:46
- 提问者网友：鐵馬踏冰河
- 2021-03-31 09:46
∫x/sin^3dx是否可积
最佳答案
- 五星知识达人网友：夜风逐马
- 2021-03-31 11:09
∫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追问是x除以sin^3
= ∫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
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