高数求助!(二重积分),我自己算总是不对,答案是 -3π⼀2,求过程。

2025-01-21 04:46:27
推荐回答(3个)
回答1:

显然∫x *cos(x+y) dy
=∫ x d sin(x+y)
=x *sin(x+y) -∫ sin(x+y) dx
=x *sin(x+y) -cos(x+y)
代入y的上下限x和0
=x *sin(2x) -cos(2x) -x *sinx+cosx
所以得到原积分
=∫(0到π) x *sin(2x) -cos(2x) -x *sinx+cosxdx
而∫ x *sin2x dx
=∫ -x/2 d(cos2x)
= -x/2 *cos2x +∫ 1/2 *cos2x dx
= -x/2 *cos2x + 1/4 sin2x
同理∫x *sinx dx
=∫ -x d(cosx)= -x *cosx +∫cosx= -x *cosx +sinx
所以∫(0到π)x *sin(2x) -cos(2x) -x *sinx+cosxdx
= -x/2 *cos2x + 1/4 sin2x -1/2 sin2x -x *cosx +sinx -sinx
= -3x/2 *cosx -1/4 sin2x
代入x的上下限π和0
= (-3π/2) *cos2π= -3π/2
就是你要的答案

回答2:

∫[0,π]∫[0,x]xcos(x+y)dydx
=∫[0,π]xsin(x+y)[0,x]dx
=∫[0,π]x(sin2x-sinx)dx
=∫[0,π]xsin2xdx-∫[0,π]xsinxdx
=-1/2∫[0,π]xdcos2x+∫[0,π]xdcosx
=-1/2xcos2x[0,π]+1/2∫[0,π]cos2xdx+xcosx[0,π]-∫[0,π]cosxdx
=-π/2+1/4sin2x[0,π]-π-sinx[0,π]
=-3π/2

回答3:

对xsin(2x)dx积分
=(-x/2)d[cos(2x)]
=-PI/2 +积分(1/2)cos(2x)dx
=-PI/2