一道不定积分题

2024-12-22 09:18:18
推荐回答(2个)
回答1:

回答2:

fy^2cos^2y/2dy
=fy^2(cosy+1)/2dy
=1/2[fy^2cosydy+fy^2dy]
其中:fy^2cosydy
=fy^2dsiny
=y^2siny-fsinydy^2
=y^2siny-2fysinydy 其中fysinydy=-fydcosy=-[ycosy-fcosydy]=siny-ycosy+c
原式=1/2[y^2siny-2siny+2ycosy+y^3/3]+C