{ (sinx cosx) / [ (a^2)((sinx)^2)+(b^2)((cosx)^2)) ]^1/2 }dx 的不定积分

这位兄弟,你也太小气了吧?你提的问题这么复杂,居然没有悬赏分?!
{ (sinx cosx) / [ (a^2)((sinx)^2)+(b^2)((cosx)^2)) ]^1/2 }dx =-∫cosx/[ (a^2)((sinx)^2)+(b^2)((cosx)^2)) ]^1/2}d(cosx)=-∫t/[a^2*(1-t^2)+b^2*t^2]^1/2 dt
(令cosx=t) =-1/2∫d((t^2)/[a^2+t^2*(b^2-a^2)]
(再令t^2=z) =-1/2∫dz/[a^2+(b^2-a^2)z]
=-1/[2(b^2-a^2)] * ∫d(b^2-a^2)z/[a^2+(b^2-a^2)z]
= -1/[2(b^2-a^2)] ln[a^2+(b^2-a^2)z] +C
=-1/[2(b^2-a^2)]ln[a^2+(b^2-a^2)*(cosx)^2]+C
对的话,不要忘记给我加分啊!