Question

高等数学，定积分

请问红线的位置是怎么求解出来的，为什么我求解出来不一样呢？

Answer 1

38(1) y^2 = 2x, x = y^2/2, dx/dy = y, s = ∫√[1+(dx/dy)^2] dy = ∫√(1+y^2)dy = [y√(1+y^2)] - ∫[y^2/√(1+y^2)]dy = 2√2 - ∫√(1+y^2)dy + ∫[1/√(1+y^2)]dy, 2s = 2√2 + ∫[1/√(1+y^2)]dy (y = tanu) = 2√2 + ∫4, π/4>secudu = 2√2 + [ln(secu+tanu)]4, π/4> = 2√2 + 2ln(1+√2), 则 s = √2 + ln(1+√2)。（2） y = lnx, y' = 1/x, s = ∫√[1+(1/x)^2] dx =∫√(1+x^2)dx/x (x = tanu) = ∫3, arctan√8>du/[sinu(cosu)^2] = -∫3, arctan√8>dcosu/[(sinu)^2(cosu)^2] = - ∫3, arctan√8>dcosu/{[1-(cosu)^2](cosu)^2} = - ∫3, arctan√8>{1/(cosu)^2 + (1/2)[1/(1-cosu)+1/(1+cosu)]}dcosu = - [-1/cosu + (1/2)ln{(1+cosu)/(1-cosu)}]3, arctan√8> = 1+(1/2)(ln3-ln2). (3) y = ∫2, x>√costdt , y' = √cosx, s = ∫2, π/2> √(1+cosx)dx = ∫2, π/2> √2cos(x/2)dx = 2√2[sin(x/2)]2, π/2> = 4.

追问

？