∫sinxcos2xdx=∫sinx(2cos²x-1)dx=-∫(2cos²x-1)d(cosx)=-(2/3)cos³x+cosx+C∫x (x-1)^1/2dx令(x-1)^1/2=u，x=u²+1，dx=2udu=∫(u²+1)u*2udu=2∫(u^4+u²)du=(2/5)u^5+(2/3)u³+C=(2/5)(x-1)^(5/2)+(2/3)(x-1)^(3/2)+C
