∫e^sin x/(e^sin x+e^cos x)dx在0~π/2上的积分

1个回答

  • 令x = π/2 - t,dx = - dt

    当x = 0,t = π/2,当x = π/2,t = 0

    L = ∫(0-->π/2) e^sinx/(e^sinx + e^cosx) dx

    = ∫(π/2-->0) e^sin(π/2 - t)/[e^sin(π/2 - t) + e^cos(π/2 - t)] · - dt

    = ∫(0-->π/2) e^cost/(e^cost + e^sint) dt

    = ∫(0-->π/2) e^cosx/(e^sinx + e^cosx) dx

    2L = ∫(0-->π/2) e^sinx/(e^sinx + e^cosx) dx + ∫(0-->π/2) e^cosx/(e^sinx + e^cosx) dx

    2L = ∫(0-->π/2) (e^sinx + e^cosx)/(e^sinx + e^cosx) dx = ∫(0-->π/2) dx = π/2

    L = ∫(0-->π/2) e^sinx/(e^sinx + e^cosx) dx = π/4