sinx/(1+sinx+cosx)在0~π/2的定积分

设t=tan(x/2)
原式=∫[0,1]2t/(1+t^2)*1/[1+2t/(1+t^2)+(1-t^2)/(1+t^2)]*2(1+t^2)dt
=∫[0,1]2t(1+t^2)/(1+t)*dt
=∫[0,1](2t^2-2t+4-4/(1+t))dt
=(2/3*t^3-t^2+4t-4ln(1+t))|[0,1]
=11/3-4ln2