(x为0到4)cos(x^1/2-1)dx的定积分要有一定的步骤!

网友回答

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