Question

希望大神指点！求不定积分1/(sinx+cosx)

Answer 1

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

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