万能代换t＝tan(x/2),则x＝2arctant,dx＝2dt/(1+t^2),cosx＝(1-t^2)/(1+t^2),所以
∫dx/(cosx+3)＝∫dt/(t^2+2)＝1/√2×arctan(t/√2)＋C＝1/√2×arctan(tan(x/2)/√2)＋C
万能代换t＝tan(x/2),则x＝2arctant,dx＝2dt/(1+t^2),cosx＝(1-t^2)/(1+t^2),所以
∫dx/(cosx+3)＝∫dt/(t^2+2)＝1/√2×arctan(t/√2)＋C＝1/√2×arctan(tan(x/2)/√2)＋C