求不定积分∫xcos(x^2)dx

速度。。。。。。。。。。。。
2024-10-31 06:58:57
推荐回答(3个)
回答1:

∫xcos(x^2)dx
=∫cos(x^2)(xdx)
=∫cos(x^2)(d(x^2)/2)
=(1/2)∫cos(x^2)d(x^2)
=(1/2)sin(x^2)+C

回答2:

∫xcos(x^2)dx
=1/2∫cos(x^2)dx^2
=1/2sinx^2+C

回答3:

∫xcos(x^2)dx=1/2∫cos(x^2)dx^2=1/2sin(x^2)+C