设x=1/cost t=arc cos(1/x)
dx=(sint/cos²t)dt
x*√(x²-1)=(1/cost)*sint/cost=sint/cos²t
所以∫[(1)/(x√(x^(2)-1))]dx
=∫(sint/cos²t)*/(sint/cos²t)*dt
=∫dt
=t
=arc cos(1/x) I(2, √2)
=arc cos(√2/2)-arc cos(1/2)
= π/4-π/3
=-π/12
dx=(sint/cos²t)dt
x*√(x²-1)=(1/cost)*sint/cost=sint/cos²t
所以∫[(1)/(x√(x^(2)-1))]dx
=∫(sint/cos²t)*/(sint/cos²t)*dt
=∫dt
=t
=arc cos(1/x) I(2, √2)
=arc cos(√2/2)-arc cos(1/2)
= π/4-π/3
=-π/12
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第1个回答 2011-10-10
设t=√（x^2-1),则dt=xdx/√（x^2-1),
原式=∫<1,3>dt/(1+t^2)=arctant|<1,3>=arctan3-π/4.
原式=∫<1,3>dt/(1+t^2)=arctant|<1,3>=arctan3-π/4.
第2个回答 2011-10-10
令x=sec t就行了最后就剩下dt
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