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∫cosxcos(x/2)dx
答案:2 悬赏:80
解决时间 2021-04-18 17:17
- 提问者网友:美人如花
- 2021-04-17 17:25
最佳答案
- 二级知识专家网友:樣嘚尐年
- 2021-04-17 17:47
∫ cosxcos(x/2) dx
= ∫ (1/2)[cos(x + x/2) + cos(x - x/2)] dx
= ∫ (1/2)cos(3x/2) + (1/2)cos(x/2) dx
= (1/2)(2/3)sin(3x/2) + (1/2)(2)sin(x/2) + C
= (1/3)sin(3x/2) + sin(x/2) + C
= ∫ (1/2)[cos(x + x/2) + cos(x - x/2)] dx
= ∫ (1/2)cos(3x/2) + (1/2)cos(x/2) dx
= (1/2)(2/3)sin(3x/2) + (1/2)(2)sin(x/2) + C
= (1/3)sin(3x/2) + sin(x/2) + C
全部回答
- 1楼网友:山鬼偶尔也合群
- 2021-04-17 18:38
∫cosxcos2xdx
=∫cosx[1-2(sinx)^2]dx
=∫cosxdx-2∫(sinx)^2dx
=∫cosxdx-∫(cos2x-1)dx
=∫cosxdx-∫cos2xdx+∫dx
=∫cosxdx-1/2∫cos2xd2x+∫dx
=sinx-1/2sin2x+x+c
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