求不定積分∫cosx/(a+bcosx)dx,
可用萬能公式代換,
設(shè)tan(x/2)=u,x=2arctanu,
dx=2du/(1+u^2),
cosx=(cosx/2)^2-(sinx/2)^2
=[1-(tanx/2)^2]/(secx/2)^2
=[1-(tanx/2)^2]/ [1+(tanx/2)^2]
=(1-u^2)/(1+u^2),
∫cosx/(a+bcosx)dx
=∫[(1-u^2)(2du)/(1+u^2)^2]/[a+b(1-u^2)/(1+u^2)]
=-2∫(u^2-1)du/[(1+u^2)(a+b+au^2-bu^2)],
設(shè)m=a+b,n=a-b,
原式=-2∫(1+u^2-2)du/[(1+u^2)(m+nu^2)]
=-2∫du/(m+nu^2)+4∫du/[(1+u^2)(m+nu^2)]
=-2∫du/(m+nu^2)+4∫[du/(m-n)]/(1+u^2)
+4∫[-n/(m-n)]du/(m+nu^2)
=[-2(m+n)/(m-n)]∫du/(m+nu^2) +4∫[du/(m-n)]/(1+u^2)
=-2(m+n)/[√(mn)(m-n) ]∫d[√（n/m）u]/[1+(u√n/m)^2]+4∫[du/(m-n)]/(1+u^2)
=-2(m+n)/[(m-n)√（mn）]arctan(√n/m)u+[4/(m-n)]arctanu]+C
=-2a/[b√(a^2-b^2)]arctan[√(a-b)/(a+b)]tan(x/2)+(2/b)arctan(tanx/2)+C.