令√x-1=u,则x=(u+1)²,dx=2(u+1)du
∫ cos(√x-1)dx
=2∫ (cosu)*(u+1)du
=2∫ ucosudu+2∫ cosudu
=2∫ ud(sinu)+2sinu
=2usinu-2∫ (sinu)du+2sinu
=2usinu+2cosu+2sinu+C
=2(√x-1)sin(√x-1)+2cos(√x-1)+2sin(√x-1)+C
则原式=(2sin1+2cos1+2sin1)-(2sin1-2cos1-2sin1)=4cos1+4sin1答案是4sin1啊。。。有个地方写错了，cos-1应该等于cos1，最后一句改成
则原式=(2sin1+2cos1+2sin1)-(2sin1+2cos1-2sin1)=4sin1