求积分∫[1/(3+sinx)]dx

问题描述:

求积分∫[1/(3+sinx)]dx
1个回答 分类:综合 2014-10-08

问题解答:

我来补答
u = tan(x/2)、dx = 2/(1 + u²) du、sinx = 2u/(1 + u²)
∫ 1/(3 + sinx) dx
= ∫ 1/[3 + 2u/(1 + u²)] * 2/(1 + u²) du
= ∫ (1 + u²)/[3(1 + u²) + 2u] * 2/(1 + u²) du
= 2∫ 1/(3u² + 2u + 3) du
= 2∫ 1/[3(u + 1/3)² + 8/3] du
= (2/3)∫ 1/[(u + 1/3)² + 8/9] du
= (2/3) * √(9/8) * arctan[(u + 1/3) * √(9/8)] + C
= (2/3) * 3/(2√2) * arctan[(3u + 1)/(2√2)] + C
= (1/√2)arctan[(3tan(x/2) + 1)/(2√2)] + C
= (1/√2)arctan[(3sinx + cosx + 1)/(2√2 + 2√2 cosx)] + C
 
 
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