大师作文网求积分∫[1/(3+sinx）］dx
来源：学生作业帮 编辑：大师作文网作业帮 分类：综合作业 时间：2024/11/12 05:48:48
求积分∫[1/(3+sinx）］dx
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
∫ 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