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

2个回答

  • ∫(- 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)