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求函数arcsinx=∫x/√(1- x^2) dx的导数?
- 答:即 y'=1/cosy=1/√[1-(siny)^2]=1/√(1-x^2)2、解题思路 分部积分法 ∫ arcsinx dx = x arcsinx - ∫ x darcsinx = x arcsinx - ∫ [x/√(1-x^2)] dx = x arcsinx + (1/2) ∫ [1/√(1-x^2)] d(1-x^2)= x arcsinx + √(1-x^2) +C ...
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2023-12-17
回答者: 题霸
1个回答
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怎样计算∫x^2/√(1- x^2) dx
- 答:解:∫xarcsinxdx =1/2*∫arcsinxdx^2 =1/2*x^2*arcsinx-1/2∫x^2darcsinx =1/2*x^2*arcsinx-1/2∫x^2/√(1-x^2)dx 令x=sint,那么,∫x^2/√(1-x^2)dx =∫(sint)^2/costdsint =∫(sint)^2dt =∫(1-cos2t)/2dt =1/2t-1/4sin2t+C=1/2t-1/2sint*cost+C ...
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2023-12-02
回答者: 格子里兮
1个回答
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1/根号下(1- x^2)的不定积分是多少?
- 答:1/根号下(1-x^2)的不定积分是 (1/2)[arcsinx + x√(1 - x²)] + C 解:x = sinθ,dx = cosθ dθ ∫ √(1 - x²) dx = ∫ √(1 - sin²θ)(cosθ dθ) = ∫ cos²θ dθ = ∫ (1 + cos2θ)/2 dθ = θ/2 + (sin2θ)/4 + C...
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2023-12-14
回答者: 所示无恒
1个回答
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不定积分根号下1- x^2怎么积分呢?
- 答:解:∫√(1-x^2)dx 令x=sint,那么 ∫√(1-x^2)dx=∫√(1-(sint)^2)dsint =∫cost*costdt =1/2*∫(1+cos2t)dt =1/2*∫1dt+1/2*∫cos2tdt =t/2+1/4*sin2t+C 又sint=x,那么t=arcsinx,sin2t=2sintcost=2x*√(1-x^2)所以∫√(1-x^2)dx=t/2+1/4*sin2t...
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2023-11-05
回答者: 寂寞的枫叶521
1个回答
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怎样求∫x arcsinx dx?
- 答:2、再换元法:令 x = sina,则可得dx = cosa da∫ x^2 ( 1/(1-x^2)^(1/2) ) dx= ∫ (sina)^2 da= ∫ (1-cos2a)/2 da= a/2 - sin2a/4= arcsinx/2 + x(1-x^2)^(1/2)/2+C3、综上所述:∫x arcsinx dx= (1/2)x^2 arcsinx - (1/2) ∫ x^2 ( 1/...
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2023-11-27
回答者: 题霸
1个回答
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如何求x^2 arcsinx dx的值域
- 答:t * (sint)^3 /3 - (1/3) ∫ (u^2-1) du = t * (sint)^3 /3 - (1/3) ( u^3 /3 - u) + C = x^3 arcsinx /3 - (1/9) (1-x^2)^(3/2) + (1/3) (1-x^2)^(1/2) + C = x^3 arcsinx /3 +(1/9) (1-x^2)^(1/2) (2+x^2) + C ...
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2024-01-01
回答者: 729707767
1个回答
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y=√(1- x^2)的导数是什么?
- 答:arcsinx的导数1/√(1-x^2)。解答过程如下:此为隐函数求导,令y=arcsinx 通过转变可得:y=arcsinx,那么siny=x。两边进行求zhuan导:cosy × y'=1。即:y'=1/cosy=1/√[1-(siny)^2]=1/√(1-x^2)
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2023-08-02
回答者: 冬雨与雪
1个回答
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根号下1- x^2的积分为多少?
- 答:根号下1-x^2的积分为1/2*arcsinx+1/2*x*√(1-x^2)+C。解:∫√(1-x^2)dx 令x=sint,那么 ∫√(1-x^2)dx=∫√(1-(sint)^2)dsint =∫cost*costdt =1/2*∫(1+cos2t)dt =1/2*∫1dt+1/2*∫cos2tdt =t/2+1/4*sin2t+C ...
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2023-11-22
回答者: zhbzwb88
3个回答
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∫arccosxdx/√(1- x^2)的极值点是?
- 答:- (2/3)∫√(1-x^2 ) dx =-x^2.√(1-x^2 ).arccosx - (1/3)x^3 -(2/3)(1-x^2 )^(3/2).arccosx - (1/3)[ arcsinx+ x/(1+x^2) ] + C // let x=sinu dx =cosu du ∫√(1-x^2 ) dx =∫ (cosu)^2 du =(1/2)∫ (1+cos2u) du =(1/2)[...
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2023-12-02
回答者: tllau38
1个回答
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为什么∫√(arcsin)(1- x^2) dx= C
- 答:结果是 (1/2)[arcsinx + x√(1 - x²)] + C x = sinθ,dx = cosθ dθ∫ √(1 - x²) dx = ∫ √(1 - sin²θ)(cosθ dθ) = ∫ cos²θ dθ= ∫ (1 + cos2θ)/2 dθ = θ/2 + (sin2θ)/4 + C= (arcsinx)/2 + (sinθcosθ...
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2023-12-17
回答者: nice千年杀
1个回答