求不定积分1/(sinx+cosx)

1个回答

  • u = tan(x / 2),dx = 2du / (1+u²)

    sinx = 2u / (1+u²),cosx = (1 - u²) / (1 + u²)

    ∫ dx / (sinx + cosx)

    = ∫ 2 / { (1 + u²) * [2u / (1+u²) + (1 - u²) / (1 + u²)] } du

    = 2∫ du / (-u² + 2u + 1)

    = 2∫ du / [2 - (u - 1)²]

    = 2∫ dy / (2 - y²),y=u - 1

    = (1 / 2√2)ln|(y + √2) / (y - √2)| + C

    = (1 / 2√2)ln|(u - 1 + √2) / (y - 1 - √2)| + C

    = (1 / 2√2)ln|[tan(x / 2) - 1 + √2] / [tan(x / 2) - 1 - √2)| + C

    = √2arctan[[tan(x / 2) - 1] / √2+ C