令 t=tan(x/2)
則cosx=[cos²(x/2)-sin²(x/2)]/[cos²(x/2)+sin²(x/2)]
=[1-tan²(x/2)]/[1+tan²(x/2)]
=(1-t²)/(1+t²)
而dx=d(2arctant)=2dt/(1+t²)
所以
∫2/(2+cosx)dx
=∫2/[2+(1-t²)/(1+t²)]*[2dt/(1+t²)]
=∫4dt/(3+t²)
=4/√3∫d(t/√3)/[1+(t/√3)²]
=4/√3arctan(t/√3)+C
代回t=tan(x/2)
故
∫2/(2+cosx)dx
=4/√3 *arctan[1/√3 *tan(x/2)] +C,C為常數(shù)