定积分根号下2-x^2的原函数怎么求?上限是根号2,下限是1

定积分根号下2-x^2的原函数怎么求?上限是根号2,下限是1
定积分根号下2-x^2的原函数怎么求?上限是根号2,下限是1

定积分根号下2-x^2的原函数怎么求?上限是根号2,下限是1
令x=√2sint
则原式=∫(π/4→π/2)√2cost*√2costdt
=∫(π/4→π/2)2cos^2(t)dt
=∫(π/4→π/2)(cos(2t)+1)dt
=sin(2t)/2|(π/4→π/2)+t|(π/4→π/2)
=0-1/2+π/4=π/4-1/2

∫(1->√2)√(2-x^2)dx
令x=√2cost
√(2-x^2)=√2sint
dx=-√2sintdt
先不忙管上下限
∫√2sint*-√2sintdt
=∫(-2sin^2t) dt
=∫(cos2t-1)dt
=∫cos2tdt-∫dt
=sin2t /2 -t
=sintcost-t

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∫(1->√2)√(2-x^2)dx
令x=√2cost
√(2-x^2)=√2sint
dx=-√2sintdt
先不忙管上下限
∫√2sint*-√2sintdt
=∫(-2sin^2t) dt
=∫(cos2t-1)dt
=∫cos2tdt-∫dt
=sin2t /2 -t
=sintcost-t
=1/2*√2sint*√2cost -t
=1/2*√(2-x^2)*x-arccos(x/√2)
∴∫(1->√2)√(2-x^2)dx
=1/2*√(2-x^2)*x-arccos(x/√2) |(1->√2)
=[1/2*√(2-2)*√2-arccos(√2/√2)]-[1/2*√(2-1)*1-arccos(1/√2)]
=(0-0)-[1/2-π/4]
=π/4-1/2

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