Question

高数定积分 求弧长

Answer 1

I = ...... = ∫<3/4, 4/3>√(1+θ^2)dθ/θ^2
令 θ = tanu， 则
I = ∫<arctan(3/4), arctan(4/3)>(secu)^3du/(tanu)^2
= ∫<arctan(3/4), arctan(4/3)>du/[cosu(sinu)^2]
= ∫<arctan(3/4), arctan(4/3)>dsinu/[(cosu)^2 (sinu)^2]
= ∫<arctan(3/4), arctan(4/3)>dsinu/{[1-(sinu)^2](sinu)^2}
= ∫<arctan(3/4), arctan(4/3)>[1/(sinu)^2 + (1/2)[1/皮答念(1-sinu) + 1/(1+sinu)]dsinu
= [-cotu +(1/2)ln{(1+sinu)/燃困(1-sinu)}]<举洞arctan(3/4), arctan(4/3)>
= 7/12 + ln(3/2)

Answer 2