热心网友
回答时间:2024-04-17 22:35
令x=2sect,dx=2secttantdt,当x从2变化到4时,t从0变化到π/3
原式=∫[0,π/3]2tant/2sect*2secttantdt=∫[0,π/3]2tan²tdt
=2∫[0,π/3](1-cos²t)/cos²t*dt=2∫[0,π/3]sec²tdt-2∫[0,π/3]dt
=2tant-2t|[0,π/3]=2√3-2π/3
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热心网友
回答时间:2024-04-17 22:37
x=2secu
dx=2secu.tanu
x=2, u=0
x=4, u=π/3
∫(2->4) √(x^2-4)/x dx
=∫(0->π/3) 4(cosu)^2/ sinu
=4∫(0->π/3) [ 1-(sinu)^2]/ sinu
=4∫(0->π/3) [ cscu -sinu ]
=4[ ln|cscu -cotu| -cosu]|(0->π/3)
=4 { [ ln(2 -√3/3) - 1/2 ] + 1 }
=4 [ ln(2 -√3/3) + 1/2 ]
=2 + 4ln(2 -√3/3)
=2 + 4ln(6 -√3) - 4ln3
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