(0至π) ∫ √(1+cos2x) dx
= (0至π) ∫ √(2cos²x) dx
= (0至π) ∫ √2 |cosx| dx
= (0至π/2) ∫ √2 cosx dx + (π/2至π) ∫ -√2 cosx dx
= [ √2 sinx ]| (0至π/2) - [ √2 sinx] | (π/2至π)
= [ √2 sin(π/2) - 0 ] - [ 0 - √2 sin(π/2) ]
= 2√2 sin(π/2)
= 2√2
本文地址: http://www.goggeous.com/20241228/1/957849
文章来源:天狐定制
版权声明:除非特别标注,否则均为本站原创文章,转载时请以链接形式注明文章出处。
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2025-01-08职业培训
2024-12-28 06:36:56职业培训
2024-12-28 06:36:56职业培训
2024-12-28 06:36:55职业培训
2024-12-28 06:36:55职业培训
2024-12-28 06:36:55职业培训
2024-12-28 06:36:47职业培训
2024-12-28 06:36:46职业培训
2024-12-28 06:36:46职业培训
2024-12-28 06:36:45职业培训
2024-12-28 06:36:44职业培训
2024-12-14 00:02职业培训
2024-12-23 12:16职业培训
2024-11-27 01:46职业培训
2024-12-04 17:08职业培训
2024-11-26 05:34职业培训
2024-12-12 07:36职业培训
2024-11-25 20:02职业培训
2025-01-01 15:23职业培训
2025-01-07 16:52职业培训
2024-12-14 15:48职业培训
扫码二维码
获取最新动态