题目

定积分∫(1+xcosx)/(1+cos^2x) 上限是π/2 下限是-π/2
答案是π/根号2

提问时间：2021-02-09

答案

∫(- π/2→π/2) (1 + xcosx)/(1 + cos^2x) dx
= ∫(- π/2→π/2) dx/(1 + cos^2x) + ∫(- π/2→π/2) xcosx dx/(1 + cos^2x)
= 2∫(0→π/2) dx/(sin^2x + cos^2x + cos^2x) + 0
= 2∫(0→π/2) dx/(sin^2x + 2cos^2x)
= 2∫(0→π/2) 1/[cos^2x(tan^2x + 2)] dx
= 2∫(0→π/2) 1/(2 + tan^2x) d(tanx)、注意1/cos^2x dx = sec^2x dx = d(tanx)
= 2 * 1/√2 * arctan(tanx/√2) |(0→π/2)、凑合公式∫ dx/(a^2 + x^2) = (1/a)arctan(x/a)
= √2 * π/2
= π/√2,(根号2分之Pi ≈ 2.2214)