Question

积分（-π/2--0）x^2 cosnx dx 怎么解

Answer 1

当n=0时，
∫ [-π/2-->0] x²dx
=1/3x³ [-π/2-->0]
=π³/24
当n>0时，
∫ [-π/2-->0] x²cosnx dx
=(1/n)∫ [-π/2-->0] x² d(sinnx)
=(1/n)x²sinnx-(1/n)∫ [-π/2-->0] 2xsinnx dx
=(1/n)x²sinnx+(2/n²)∫ [-π/2-->0] x d(cosnx)
=(1/n)x²sinnx+(2/n²)xcosnx-(2/n²)∫ [-π/2-->0] cosnxdx
=(1/n)x²sinnx+(2/n²)xcosnx-(2/n³)sinnx [-π/2-->0]
=-(1/n)(π²/4)sin(-nπ/2)-(2/n²)(-π/2)cos(-nπ/2)-(2/n³)sin(-nπ/2)
=(1/n)(π²/4)sin(nπ/2)+(2/n²)(π/2)cos(nπ/2)+(2/n³)sin(nπ/2)

Answer 2

∫udv=uv-∫udv
多次使用分部积分，把x^2降次就行了。
∫x^2.cosnx dx
=1/n*∫x^2 * d(sinnx)
=1/n*(x^2*sinnx-∫sinnxd(x^2))
=1/n*(x^2*sinnx-∫2xsinnxdx)
=1/n(x^2*sinnx+2/n*∫xd(cosnx))
=x^2/n*sinnx+2/n^2*(xcosnx-∫cosnxdx)
=x^2/n*sinnx+2/n^2*xcosnx-2/n^2*1/n*(sinnx)+C
=1/n^3(n^2*x^2*sinnx+2nxcosnx-2sinnx)+C
再代入求值就行了

Answer 3

把cosnx凑到dx里面，分步积分