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已知函数f(x)= asint(a^2- x^2),求
- 答:a^2-x^2 =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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2024-05-20
回答者: 风林网络手游平台
1个回答
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∫( cos2t+1)/2dt的积分表达式是什么
- 答:a^2-x^2 =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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2024-05-25
回答者: 风林网络手游平台
1个回答
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∫(sin2t+1)/2ds=?
- 答:a^2-x^2 =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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2024-05-25
回答者: 风林网络手游平台
1个回答
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∫√( a^2- x^2) dx=什么?
- 答:a^2-x^2 =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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已知函数f(x)= asint^2,求积分
- 答:a^2-x^2 =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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已知a^2- x^2=1,求∫√( a^2- x^2) dx。
- 答:a^2-x^2 =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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求不定积分∫√( a^2- x^2)的值域?
- 答:a^2-x^2 =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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2024-05-20
回答者: 风林网络手游平台
1个回答
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已知x= asint求积分值域是什么意思?
- 答:a^2-x^2 =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
1
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求∫(a^2- x^2) dx的值域?
- 答:a^2-x^2 =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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2024-05-30
回答者: 风林网络手游平台
1个回答
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求积分∫√(a^2- x^2) dx的值域是多少?
- 答:a^2-x^2 =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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2024-05-20
回答者: 风林网络手游平台
1个回答