①∫(0到π/4)xcos2x dx；②∫(0到e-1)(1+x)ln^2(1+x)dx
1个回答
1.原积分=∫(0到π/4)x×1/2×dsin2x
=1/2×xsin2x(x=π/4)-1/2×xsin2x(x=0)-1/2∫(0到π/4)sin2x dx
=π/8+1/4×cos2x(x=π/4)-1/4×cos2x(x=0)
=π/8-1/4.
2.原积分=∫(0到e-1)ln^2(1+x)×1/2×d(1+x)^2
=1/2×(1+x)^2×ln^2(1+x)(x=e-1)-1/2×(1+x)^2×ln^2(1+x)(x=0)-1/2∫(0到e-1)(1+x)^2dln^2(1+x)
=1/2×e^2×(1+ln2)-1/2×ln2-1/2∫(0到e-1)(1+x)^2×1/[2(1+x)]×2dx
=1/2×e^2×(1+ln2)-1/2×ln2-1/2∫(0到e-1)(1+x)dx
=1/2×e^2×(1+ln2)-1/2×ln2-1/4×(1+x)^2(x=e-1)+1/4×(1+x)^2(x=0)
=1/4×e^2+1/2×e^2×ln2-1/2×ln2+1/4.
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