设tan(x/2)=t 则sin x=2t/(1+t^2) cos x=(1-t^2)/(1+t^2) dx=2/(1+t^2)dt
∫sinx/(sinx+cosx)dx=∫2t/(1+t^2)*2/(1+t^2)/[2t/(1+t^2) +(1-t^2)/(1+t^2)]dt
=-4∫t/[(1+t^2)(t+√2-1)(t-√2-1)dt
设：t/[(1+t^2)(t+√2-1)(t-√2-1)=A/(t-√2-1)+B/(t+√2-1)+(Ct+D)/(1+t^2)
A(t+√2-1)(1+t^2)+B(t-√2-1)(1+t^2)+(Ct+D)(t+√2-1)(t-√2-1)=t
t=1-√2:-2√2(4-2√2)B=1-√2 B=1/8
t=1+√2:2√2(4+2√2)A=1+√2 A=1/8
t=0:1/8(√2-1)+1/8(-√2-1)+D(√2-1)(-√2-1)=0 D=-1/4
t=1:1/8(1+√2-1)(1+1)+1/8(1-√2-1)(1+1)+(C-1/4)(1+√2-1)(1-√2-1)=1 C=-1/4
原式=-4∫[1/8/(t-√2-1)+1/8/(t+√2-1)-1/4(t+1)/(1+t^2)]dt
=-1/2∫dt/(t-√2-1)-1/2∫dt/(t+√2-1)+∫t/(1+t^2)dt+∫dt/(1+t^2)
=-1/2ln|t-√2-1|-1/2ln|t+√2-1|+1/2ln|1+t^2|+arctan t +C
=-1/2ln[(t-√2-1)(t+√2-1)/(1+t^2)]+arctan t +C
= x/2-1/2*(ln(sinx+cosx))+C
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