方法一:
∵f(x)=x³cosx为奇函数
∴∫(-π/2, π/2) x³cosx dx=0
方法二:
∫(-π/2, π/2) x³cosx dx
=∫(-π/2, π/2) x³ d(sinx)
=x³sinx |(-π/2, π/2)-∫(-π/2, π/2) 3x²sinx dx
=∫(-π/2, π/2) 3x² d(cosx)
=3[x²cosx |(-π/2, π/2)-∫(-π/2, π/2) 2xcosx dx]
=-6∫(-π/2, π/2) xd(sinx)
=-6[xsinx |(-π/2, π/2)-∫(-π/2, π/2) sinx dx]
=-6cosx |(-π/2, π/2)
=0
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