高数定积分,求过程?
4个回答
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(5)
let
x=tany
dx=(secy)^2.dy
x=1,y=π/4
x=√3,y=π/3
∫(1->√3)dx/[x^2.√(1+x^2)]
=∫(π/4->π/3)(secy/tany)dy
=∫(π/4->π/3)cscydy
=[ln、cscy-coty、]、(π/4->π/3)
=ln、2-√3/3、-ln、√2-1、
(6)
∫(0->a)x^2.√(a^2-x^2)]
let
x=asiny
dx=acosydy
x=0,a=0
x=a,y=π/2
∫(0->a)x^2.√(a^2-x^2)]
=a^4.∫(0->π/2)(sinycosy)^2dy
=(a^4/4).∫(0->π/2)(sin2y)^2dy
=(a^4/8).∫(0->π/2)(1-cos2y)dy
=(a^4/8)[(y-(1/2)sin2y)]、(0->π/2)
=(πa^4/16)
let
x=tany
dx=(secy)^2.dy
x=1,y=π/4
x=√3,y=π/3
∫(1->√3)dx/[x^2.√(1+x^2)]
=∫(π/4->π/3)(secy/tany)dy
=∫(π/4->π/3)cscydy
=[ln、cscy-coty、]、(π/4->π/3)
=ln、2-√3/3、-ln、√2-1、
(6)
∫(0->a)x^2.√(a^2-x^2)]
let
x=asiny
dx=acosydy
x=0,a=0
x=a,y=π/2
∫(0->a)x^2.√(a^2-x^2)]
=a^4.∫(0->π/2)(sinycosy)^2dy
=(a^4/4).∫(0->π/2)(sin2y)^2dy
=(a^4/8).∫(0->π/2)(1-cos2y)dy
=(a^4/8)[(y-(1/2)sin2y)]、(0->π/2)
=(πa^4/16)
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2021-11-01
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详细过程如图rt所示……希望能帮到你
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