令u=tan(x/2)，则sinx=2u/(1+u^2)，dx=2/(1+u^2)du原式=∫1/[3+2u/(1+u^2)]*2du/(1+u^2)=∫2du/(3u^2+2u+3)=(2/3)*∫du/[(u+1/3)^2+8/9]=(2/3)*(3/2√2)*arctan[(3u+1)/2√2]+C=(1/√2)*arctan[(3/2√2)*tan(x/2)+1/2√2]+C，其中C是任意常数
