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d/dx∫sin(x-t)^2dt 积分上限x下限0
人气:336 ℃ 时间:2020-06-04 14:02:41
解答
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
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