1/(cosa+sina)对a积分怎么做?

Question

1/(cosa+sina)对a积分怎么做?

Answer 1

令 tan(a/2)=x,则 sina=2x/(1+x^2),cosa=(1-x^2)/(1+x^2),da=2dx/(1+x^2),得
I = ∫ da/(cosa+sina) = ∫ 2dx/(1+2x-x^2) = -∫ 2dx/[(x-1)^2-2]
= ∫ -2dx/[(x-1)^2-2] = (1/√2)∫{1/(x+√2-1)-1/[x-(√2+1)]}dx
= (1/√2)[ln|x+√2-1| - ln|x-(√2+1)| ]+C
= (1/√2)[ln|(x+√2-1) /[x-(√2+1)]| +C
= (1/√2)[ln|(tan(a/2)+√2-1) /[tan(a/2)-(√2+1)]| +C