求定积分∫上限π/4,下限0 tan^2xdx,
问题描述：
求定积分∫上限π/4,下限0 tan^2xdx,
答
原式=∫(0,π/4)(sec²x-1)dx
=∫(0,π/4)sec²xdx-∫(0,π/4)dx
=∫(0,π/4)d(tanx)-∫(0,π/4)dx
=(tanx-x)|(0,π/4)
=1-π/4