急啊啊,谁知道这一个怎么求不定积分 根号下x-x^2的不定积分,用换元法啊啊
答案:2 悬赏:20
解决时间 2021-02-01 22:51
- 提问者网友:独菊痴梦
- 2021-02-01 12:27
急啊啊,谁知道这一个怎么求不定积分 根号下x-x^2的不定积分,用换元法啊啊
最佳答案
- 二级知识专家网友:厌今念往
- 2021-02-01 13:41
8)arcsin(2x - 1) + (1/2)^2sin^2(y)] = (1/2)cosy
则∫ √(x - x^2) dx
= ∫ [(1/2)cosy]^2 dy
= (1/2)siny、dx = (1/2)cosy dy
siny = 2x - 1;2)^2 - (1/8)y + (1/8)sinycosy + C
= (1/√(x - x^2) = √[(1/2)^2 - (x - 1/2)^2]
令x - 1/2 = (1/8)∫ (1 + cos2y) dy
= (1/8)[y + (1/2)sin2y] + C
= (1/4)∫ cos^2(y) dy
= (1/、cos^2(y) = 1 - (2x - 1)^2 = 1 - (4x^2 - 4x + 1) = 4x - 4x^2 → cosy = 2√(x - x^2)
√(x - x^2) = √[(1/
则∫ √(x - x^2) dx
= ∫ [(1/2)cosy]^2 dy
= (1/2)siny、dx = (1/2)cosy dy
siny = 2x - 1;2)^2 - (1/8)y + (1/8)sinycosy + C
= (1/√(x - x^2) = √[(1/2)^2 - (x - 1/2)^2]
令x - 1/2 = (1/8)∫ (1 + cos2y) dy
= (1/8)[y + (1/2)sin2y] + C
= (1/4)∫ cos^2(y) dy
= (1/、cos^2(y) = 1 - (2x - 1)^2 = 1 - (4x^2 - 4x + 1) = 4x - 4x^2 → cosy = 2√(x - x^2)
√(x - x^2) = √[(1/
全部回答
- 1楼网友:闲懒诗人
- 2021-02-01 14:17
令√(x^2-9)=u,则:x^2=u^2+9,∴d(x^2)=2udu。
∴∫[√(x^2-9)/x]dx
=(1/2)∫[2x√(x^2-9)/x^2]dx
=(1/2)∫[√(x^2-9)/x^2]d(x^2)
=(1/2)∫[u/(u^2+9)]·2udu
=∫{[(u^2+9)-9]/(u^2+9)}du
=∫du-9∫[1/(u^2+9)]du
=u-9∫{1/[9(u/3)^2+9]}du
=u-3∫{1/[(u/3)^2+1]}d(u/3)
=u-3arctan(u/3)+c
=√(x^2-9)-3arctan[(1/3)√(x^2-9)]+c。
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