高等数学:求定积分如图所示该怎么解答,有多少种做法啊
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设θ=tanx
0<=θ<=2π
0<=x<=arctan(2π)
tanx=2π =>
1+θ^2=(secx)^2
dθ=(secx)^2dx
0->2π ∫sqrt(1+θ^2)dθ=
0->2π ∫ (secx)^3dx=
0->2π ∫ (tanx)^2*secx dx +0->2π ∫ secx dx=
=1/2 (tan(x) sec(x)-log(cos(x/2)-sin(x/2))+log(sin(x/2)+cos(x/2))) | 0->arctan(2π)
π*sqrt(1+4*π^2)+1/2 sinh^(-1)(2 pi)≈21.2563
0<=θ<=2π
0<=x<=arctan(2π)
tanx=2π =>
1+θ^2=(secx)^2
dθ=(secx)^2dx
0->2π ∫sqrt(1+θ^2)dθ=
0->2π ∫ (secx)^3dx=
0->2π ∫ (tanx)^2*secx dx +0->2π ∫ secx dx=
=1/2 (tan(x) sec(x)-log(cos(x/2)-sin(x/2))+log(sin(x/2)+cos(x/2))) | 0->arctan(2π)
π*sqrt(1+4*π^2)+1/2 sinh^(-1)(2 pi)≈21.2563
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