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∫√( a^2- x^2) dx的积分怎么算啊?
- 答：=a^2-a^2sint^2 =a^2cost^2 ∫√（a^2-x^2）dx =∫acost*acostdt =a^2∫cost^2dt =a^2∫（cos2t+1）/2dt =a^2/4∫(cos2t+1)d2t =a^2/4*(sin2t+2t)将x=asint代回，得：∫√（a^2-x^2）dx=x√(a^2-x^2)/2+a^2*arcsin(x/a)/2+C（C为常数）...
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2024-06-06
回答者: 风林网络手游平台
1个回答
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f(x)=sinx/2*cosx/2+cos^2(x/2)-2求函数f(x)在[π,17/12π]上的最大...
- 问：f(x)=sinx/2*cosx/2+cos^2(x/2)-2求函数f(x)在[π,17/12π]上的最大值和最...
- 答：f(x)=sinx/2*cosx/2+cos^2(x/2)-2 =1/2sinx+1/2(cosx+1)-2 =1/2(sinx+cosx)=√2/2sin(x+π/4)y=sinX在(π,3π/2)是减函数 π<x+π/4<3π/2, 3π/4<x<5π/4 y=√2/2sin(x+π/4)在3π/4<x<5π/4是减函数 最大值x=π , y=-1/2 最小值x=5π/4...
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2010-01-25
回答者: 缺衣少食2468
2个回答
3
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∫√( a^2- x^2) dx=什么?
- 答：=a^2-a^2sint^2 =a^2cost^2 ∫√（a^2-x^2）dx =∫acost*acostdt =a^2∫cost^2dt =a^2∫（cos2t+1）/2dt =a^2/4∫(cos2t+1)d2t =a^2/4*(sin2t+2t)将x=asint代回，得：∫√（a^2-x^2）dx=x√(a^2-x^2)/2+a^2*arcsin(x/a)/2+C（C为常数）...
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2024-06-10
回答者: 风林网络手游平台
1个回答
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∫√(a^2- x^2) dx怎么积分
- 答：=a^2-a^2sint^2 =a^2cost^2 ∫√（a^2-x^2）dx =∫acost*acostdt =a^2∫cost^2dt =a^2∫（cos2t+1）/2dt =a^2/4∫(cos2t+1)d2t =a^2/4*(sin2t+2t)将x=asint代回，得：∫√（a^2-x^2）dx=x√(a^2-x^2)/2+a^2*arcsin(x/a)/2+C（C为常数）...
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2024-06-14
回答者: 风林网络手游平台
1个回答
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求过程:cosx-sinx=2^(1/2)[cos(x+π/4)]
- 问：谁能给出上式推演的具体过程？ 说明：2^(1/2)就是根2
- 答：一般是从右边证到左边，应用和差化积公式即可。2^(1/2)[cos(x+π/4)]=2^(1/2)cosxcosπ/4-2^(1/2)sinxsinπ/4=2^(1/2)(1/2)^(1/2)cosx-2^(1/2)(1/2)^(1/2)sinx=cosx-sinx.
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2007-06-21
回答者: quesaisje
2个回答
