令cosx=a(cosx+sinx)+b(cosx+sinx)'=(a+b)cosx+(a-b)sinx===>a=b=1/2∫cosx / (cosx+sinx)dx =(1/2)∫[(cosx+sinx)+(cosx+sinx)'] / (cosx+sinx)dx =(1/2)[x+ln|cosx+sinx|]+C