Question

不定积分求高手解答

第八小题我怎么算都和标准答案不对，求高手帮忙写一下过程感谢

Answer 1

使用分部积分法。
设 u = x², dv = cos(2x)dx。则有 du = 2x *dx，v = 1/2 * sin(2x)。那么，原积分：
= ∫u*dv
= u * dv - ∫v * du
= 1/2 * sin(2x) * x²|x=0→π - ∫[1/2 * sin(2x) * 2x * dx]
= 1/2 *[π²*sin(2π) - 0² *sin0] - ∫x * sin(2x) *dx
= -∫x * sin(2x) *dx
再设 r = x，ds = sin(2x)*dx。则 dr = dx, s = -1/2 * cos(2x)。那么：
∫x * sin(2x) * dx
=∫r *ds
= r * s - ∫s * dr
= -1/2 * x * cos(2x)|x=0→π + 1/2 *∫cos(2x) *dx
=-1/2 * [π *cos(2π) - 0 *cos0] + 1/4 * sin(2x)|x=0→π
=-1/2 * π + 1/4 * [sin(2π) - sin0]
=-π/2
所以，原积分：
= π/2

Answer 2

∫x²cos2xdx

=1/2*∫x²dsin2x
=1/2*x²sin2x-1/2*∫2xsin2xdx
=1/2*x²sin2x+1/2*∫xdcos2x
=1/2*x²sin2x+1/2*xcos2x-1/2*∫cos2xdx
=1/2*x²sin2x+1/2*xcos2x-1/4*sin2x+C
你的1/2系数全部不见了???

Answer 3

第一步就错了， d(1/2*sin2x)=cos2xdx
∫[0,π]x²cos2xdx
=1/2*∫[0,π]x²dsin2x
=1/2*x²sin2x|[0,π]-1/2*∫[0,π]sin2x*2xdx
=0+1/2*∫[0,π]xdcos2x
=1/2*xcos2x|[0,π]-1/2*∫[0,π]cos2xdx
=π/2-1/4*sin2x|[0,π]
=π/2-0
=π/2