第1个回答 2014-12-02
∵f(x)在[0,a]连续 ∴f(x)在[0,a]可积
令 x=a-t 则dx=-dt
∫[0,a]f(x)dx=-∫[a,0]f(a-t)dt
=∫[0,a]f(a-t)dt
=∫[0,a]f(a-x)dx (定积分与积分变量的选择无关)
∫[0,π/4](1-sin2x)/(1+sin2x)
=∫[0,π/4][1-sin2(π/4-x)]/[1+sin2(π/4-x)]
=∫[0,π/4][1-sin(π/2-2x)]/[1+sin(π/2-2x)]
=∫[0,π/4][1-cos(2x)]/[1+cos(2x)]
=∫[0,π/4][2sin^2(x)]/[2cos^2(x)]
=∫[0,π/4][1-cos^2(x)]/cos^2(x)
=[tanx-x][0,π/4]
=1-π/4