求∫(x,0) e^t^2dt⼀∫(d,0) e^2t^2dt极限

求∫(x,0) e^t^2dt/∫(d,0) e^2t^2dt极限
2024-11-20 08:23:16
推荐回答(1个)
回答1:

先求∫(0->x)sin(x-t)^2dt
=∫(0->x)(1-cos(2x-2t)/2 dt
=1/2∫(0->x)dt-1/2∫(0->x)cos(2x-2t)dt
=x/2+1/4∫(0->x)cos(2x-2t)d(2x-2t)
=x/2+1/4sin(2x-2t)|(0->x)
=x/2+1/4(sin(2x-2x)-sin(2x-2*0)
=x/2+sin2x/4

所以
d/dx∫(0->x)sin(x-t)^2dt
=d(x/2+sin2x/4)/dx
=1/2+1/4*cos2x*2
=1/2+cos2x /2