Question

帮忙写一下过程

Answer 1

∫(-π/4,π/4)tanxdx
=∫(-π/4,π/4)sinx/cosxdx
=∫(-π/4,π/4)-1/cosxd(cosx)
=-lncosx|(-π/4,π/4)
=-[lncos(π/4)-lncos(-π/4)]
=-[ln(√2/2)-ln(√2/2)]
=0
∫(-π,π)sin^3xdx
=-∫(-π,π)sin^2xd(cosx)
=-∫(-π,π)(1-cos^2x)d(cosx)
=-[cosx-1/3cos^3x]|(-π,π)
=[1/3cos^3x-cosx]|(-π,π)
=(1/3cos^3π-cosπ)-[1/3cos^3(-π)-cos(-π)]
=1/3cos^3π-cosπ-1/3cos^3π+cosπ
=0

Answer 2

第一个定积分的几何含义是：tanX曲线在-π/4到π/4的区间与X轴围成的面积
第二个定积分的几何含义是：sin³X曲线在-π到π的区间与X轴围成的面积