设A=∫cosx/(2sinx+3cosx)dx,B=∫sinx/(2sinx+3cosx)dx,则
3A+2B=3∫cosx/(2sinx+3cosx)dx+2∫sinx/(2sinx+3cosx)dx=∫dx=x+C1
2A-3B=∫2cosx/(2sinx+3cosx)dx-∫3sinx/(2sinx+3cosx)dx=∫(2cosx-3sinx)/(2sinx+3cosx)dx
=∫d[(2sinx+3cosx)]/[(2sinx+3cosx)]=ln|2sinx+3cosx|+C2
解得
A=1/13*[(3x+3C1+2ln|2sinx+3cosx|+2C2]=1/13*[(3x+2ln|2sinx+3cosx|+3C1+2C2]=
1/13*(3x+2ln|2sinx+3cosx|)+C