最后一步跌倒
u=tan(x/2),sinx=2u/1+u^2,du=[1+[tan(x/2)]^2]/2dx=((1+u^2)/2)dx
dx=2du/(1+u^2)
3+(sinx)^2=(3+3u^4+6u^2+4u^2)/(1+u^2)^2
=(3u^4+10u^2+3)/(1+u^2)^2
∫dx/[3+(sinx)^2]=2∫(1+u^2)du/(3u^4+10u^2+3)
=2∫(1+u^2)du/[(3u^2+1)(u^2+3)]
=(1/2)∫[(3u^2+1)+(u^2+3)]du/[(3u^2+1)(u^2+3)]
=(1/2)∫du/(3u^2+1)+(1/2)∫du/(u^2+3)
=(1/(2√3)∫d(√3u)/[(√3u)^2+1] +(√3/6)∫d(u/√3)/[(u/√3)^2+1]
=(1/(2√3))arctan(√3u+1)+(√3/6)arctan(u/√3)
=(1/(2√3))arctan[√3tan(x/2)+1]+(√3/6)arctan[tan(x/2)/√3]+C