第1个回答 2012-06-10
∫[0,2π] cosθdθ/(l+rcosθ)
=(1/r)∫[0,2π](rcosθ+l)dθ/(rcosθ+l)-(1/r)∫[0,2π]ldθ/(l+rcosθ)
=(1/r)2π -(1/r)∫[0,2π]dθ/[1+(r/l)(2cos(θ/2)^2-1)]
=2π/r -(1/r)∫[0,2π]dtan(θ/2)/[((1-r/l)/2)tan(θ/2)^2+(1-r/l)/2+r/l]
=2π/r-(1/r)∫[0,2π]dtan(θ/2)[((1-r/l)/2)tan(θ/2)^2+(1+r/l)/2]
=2π/r -(1/r)[1/√[(1-r/l)(1+r/l)]∫[0,2π] [d√(1-r/l)/(1+r/l)tan(θ/2) /[(1-r/l)/(1+r/l)tan(θ/2)^2+1]
=2π/r -(1/r)[1/√(1-r^2/l^2)]arctan[√[(1-r/l)/(1+r/l)]tanθ/2 ] |[0,2π]
=2π/r
∫dx/(1+cosx)=∫d(x/2)/cos(x/2)^2=∫dtan(x/2)本回答被网友采纳
=(1/r)∫[0,2π](rcosθ+l)dθ/(rcosθ+l)-(1/r)∫[0,2π]ldθ/(l+rcosθ)
=(1/r)2π -(1/r)∫[0,2π]dθ/[1+(r/l)(2cos(θ/2)^2-1)]
=2π/r -(1/r)∫[0,2π]dtan(θ/2)/[((1-r/l)/2)tan(θ/2)^2+(1-r/l)/2+r/l]
=2π/r-(1/r)∫[0,2π]dtan(θ/2)[((1-r/l)/2)tan(θ/2)^2+(1+r/l)/2]
=2π/r -(1/r)[1/√[(1-r/l)(1+r/l)]∫[0,2π] [d√(1-r/l)/(1+r/l)tan(θ/2) /[(1-r/l)/(1+r/l)tan(θ/2)^2+1]
=2π/r -(1/r)[1/√(1-r^2/l^2)]arctan[√[(1-r/l)/(1+r/l)]tanθ/2 ] |[0,2π]
=2π/r
∫dx/(1+cosx)=∫d(x/2)/cos(x/2)^2=∫dtan(x/2)本回答被网友采纳
第2个回答 2012-06-09
三角换元啊，tan(theta/2)=t。化成有理分式的积分，然后原函数都能算出来啊追问
使用万能公式吗？然后是把tan²α设为t，还是tanα=t，还是积不出啊，谢谢
追答如果你遇到一般的有理分式的积分，方法是类似的，就是把它分解为若干个分母为1次或者2次多项式的和，然后逐项积分。
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