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求定积分 根号(1-x^2)/x^2dx 上1 下根号2/2
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let
x= sinu
dx= cosu du
x=√2/2 , u=π/4
x=1, u=π/2
∫(√2/2->1) √(1-x^2) /x^2 dx
=∫(π/4->π/2) [cosu /(sinu)^2] [ cosu du]
=∫(π/4->π/2) (cotu)^2 du
=∫(π/4->π/2) [(cscu)^2 -1 ] du
=[-cotu -u ]|(π/4->π/2)
= (0 - π/2 ) - ( -1 -π/4)
=1 - π/4
x= sinu
dx= cosu du
x=√2/2 , u=π/4
x=1, u=π/2
∫(√2/2->1) √(1-x^2) /x^2 dx
=∫(π/4->π/2) [cosu /(sinu)^2] [ cosu du]
=∫(π/4->π/2) (cotu)^2 du
=∫(π/4->π/2) [(cscu)^2 -1 ] du
=[-cotu -u ]|(π/4->π/2)
= (0 - π/2 ) - ( -1 -π/4)
=1 - π/4
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