当limx→0 求∫(0,x)[∫(0,u^2)arctan(1+t)dt]du/x(1-cosx)

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当limx→0 求∫(0,x)[∫(0,u^2)arctan(1+t)dt]du/x(1-cosx)

当limx→0 求∫(0,x)[∫(0,u^2)arctan(1+t)dt]du/x(1-cosx)
当limx→0 求∫(0,x)[∫(0,u^2)arctan(1+t)dt]du/x(1-cosx)

当limx→0 求∫(0,x)[∫(0,u^2)arctan(1+t)dt]du/x(1-cosx)
分子:∫(0,x)[∫(0,u^2)arctan(1+t)dt]du
分母:x(1-cosx) (1/2)x^3
用洛必达法则,上下同时求导
分子变为:∫(0,x^2)arctan(1+t)dt
分母变为:(3/2)x^2
继续用洛必达法则
分子变为:arctan(1+x) * 2x
分母变为:3x
所以原式=lim(x->0)(2/3)arctan(1+x)
=(2/3)arctan1
=(2/3)(π/4)
=π/6

lim(x→0) ∫(0,x)[∫(0,u^2)arctan(1+t)dt]du/[x(1-cosx)]
=lim(x→0) ∫(0,x)[∫(0,u^2)arctan(1+t)dt]du/[x^3/2] (0/0)
=lim(x→0) 2[∫(0,x^2)arctan(1+t)dt]/[3x^2] (0/0)
=lim(x→0) 2x*arctan(1+x^2)/[3x]
=2/3*π/4
=π/6