最佳答案

设y=tan(x/2)，那么dx=2dy/(1+y²)，

sinx=2y/(1+y²)，cosx=(1-y²)/(1+y²)。

∫dx/(2sinx-cosx-5)

=∫[2dy/(1+y²)]/[4y/(1+y²)-(1-y²)/(1+y²)-5]

=2∫dy/[4y+(y²-1)-5(1+y²)]

=-2∫dy/(4y²-4y+6)

=-∫d(2y-1)/[(2y-1)²+5]

=-(1/根号5)arctan[(2y-1)/根号5]+c

=-(1/根号5)arctan{[2tan(x/2)-1]/根号5}+c