borenoApple
幼苗
共回答了18个问题采纳率:100% 举报
显然,R和r不能同时为0
(1)当R=r时,原式=(1/(2r²))∫dθ/(1-cosθ)
=(1/(2r²))∫dθ/(2sin²(θ/2))
=(1/(4r²))∫csc²(θ/2)dθ
=-cot(θ/2)/(4r²)+C (C是积分常数);
(2)当R≠r时,设x=tan(θ/2),则dθ=2dx/(1+x²),cosθ=(1-x²)/(1+x²)
原式=∫(2dx/(1+x²))/(R²+r²-2rR((1-x²)/(1+x²)))
=∫2dx/((R²+r²)(1+x²)-2rR(1-x²))
=2∫dx/((R-r)²+(R+r)²x²)
=2/(R²-r²)arctan((R+r)x/(R-r))+C (C是积分常数)
=2/(R²-r²)arctan((R+r)(tan(θ/2))/(R-r))+C.
如果积分上下限是0-2π,可以自己代入求解.但是当R=r时,积分上下限是0-2π的积分不存在.
1年前
2