Question

求教不定积分(2)，希望有过程，谢谢！

求教不定积分(2)，希望有过程，谢谢！注：目前刚学到分部积分法．

Answer 1

令z = tan(x/2)，dx = 2dz/(1 + z²)，万能代换，cosx = (1 - z²)/(1 + z²)
∫ dx/(2 + cosx)
= ∫ 2/(1 + z²) * 1/[2 + (1 - z²)/(1 + z²)] dz
= ∫ 2/(1 + z²) * (1 + z²)/(2 + 2z² + 1 - z²) dz
= ∫ 2/(z² + 3) dz
= 2 * 1/√3 * arctan(z/√3) + C <== ∫ dx/(x²+a²)=(1/a)arctan(x/a)+C
= (2/√3)arctan[(1/√3)tan(x/2)] + C

Answer 2

题目呢