数学分析题目求教
1个回答
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1、原式=(1/2)*[∫(0,π/2) (sinθ+cosθ)/(sinθ+cosθ)dθ+∫(0,π/2) (cosθ-sinθ)/(sinθ+cosθ)dθ]
=(1/2)*[∫(0,π/2) dθ+∫(0,π/2) d(sinθ+cosθ)/(sinθ+cosθ)]
=(1/2)*[θ+ln|sinθ+cosθ|]|(0,π/2)
=(1/2)*(π/2)
=π/4
2、原式=-∫(0,+∞) cosxd[e^(-x)]
=-cosxe^(-x)|(0,+∞)-∫(0,+∞) sinxe^(-x)dx
=1+∫(0,+∞) sinxd[e^(-x)]
=1+sinxe^(-x)|(0,+∞)-∫(0,+∞) e^(-x)cosxdx
=1-∫(0,+∞) e^(-x)cosxdx
所以原式=1/2
3、题目有误
=(1/2)*[∫(0,π/2) dθ+∫(0,π/2) d(sinθ+cosθ)/(sinθ+cosθ)]
=(1/2)*[θ+ln|sinθ+cosθ|]|(0,π/2)
=(1/2)*(π/2)
=π/4
2、原式=-∫(0,+∞) cosxd[e^(-x)]
=-cosxe^(-x)|(0,+∞)-∫(0,+∞) sinxe^(-x)dx
=1+∫(0,+∞) sinxd[e^(-x)]
=1+sinxe^(-x)|(0,+∞)-∫(0,+∞) e^(-x)cosxdx
=1-∫(0,+∞) e^(-x)cosxdx
所以原式=1/2
3、题目有误
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