{"topic":{"id":13873,"title":"（已解决）求解不定积分","slug":"topic","categoryId":54,"createdAt":"2024-11-18 08:11:05.977763","bumpedAt":"2024-11-18 08:29:17.486384","views":180,"postsCount":7,"highestPostNumber":7,"likeCount":3,"closed":false,"pinned":false,"pinnedGlobally":false,"pinnedAt":null,"pinnedUntil":null,"isBanner":false,"excerpt":"∫1+cos⁡x1+sin⁡2x dx\\displaystyle \\int \\dfrac{1+\\cos{x}}{1+\\sin^2{x}}~\\mathrm{d}x∫1+sin2x1+cosx dx 该不定积分如何计算？看了看 ggb 和 mathematica 的结果，有些复杂，不太像大一能写的感谢门友相助，已解决（但是没找到“解决方案”按钮），下面梳理一下思路取代换 t=tan⁡(x2)t=\\tan\\left(\\dfrac{x}{…","deletedAt":null,"isAnonymous":false,"category":{"id":54,"name":"学术专区","slug":"academic","color":"9EB83B"},"author":{"id":5905,"username":"PipaQinse233","name":"PipaQinse233","avatarUrl":"/uploads/image/f7/1b/f71bbda87a32de08df517fe7ee4a14e3f63bfd1b2105d3ddb8ba1f4c2eaa1c29.avif"},"tags":[]},"posts":[{"id":43342,"postNumber":1,"raw":"$\\displaystyle \\int \\dfrac{1+\\cos{x}}{1+\\sin^2{x}}~\\mathrm{d}x$\n\n该不定积分如何计算？看了看 ggb 和 mathematica 的结果，有些复杂，不太像大一能写的\n\n[grid mode=\"tile\"]\n![ggb|281x500](/uploads/image/89/ae/89ae4faec1a1f6870a701e4d6d59b8c622173bce1b76f9f7979f6fb07eb5bd6f.avif)\n![mathematica|230x500](/uploads/image/d5/f6/d5f60ccf3ba9287b65c27049a649755b5b197c6848e1e2abae6ff23696b625d1.avif)\n[/grid]\n\n---\n\n感谢门友相助，已解决（但是没找到“解决方案”按钮），下面梳理一下思路\n\n1. 取代换 $t=\\tan\\left(\\dfrac{x}{2}\\right)$，得到一个有理函数的不定积分\n2. 对于分母中的二次三项式 $x^2+px+q$，可先配方得 $\\left(x+\\dfrac{p}{2}\\right)^2-\\dfrac{p^2}{4}+q$，再由平方差公式进一步化简，得到两个分式（打完才想起来 $x$ 用过了，懒得改了）\n3. 最后查表并化简\n\n最终结果（如果我没算错的话）：$\\dfrac{\\sqrt{2}\\left(\\sqrt{3+2\\sqrt{2}}\\arctan\\left(\\dfrac{\\tan\\dfrac{x}{2}}{\\sqrt{3-2\\sqrt{2}}}\\right)-\\sqrt{3-2\\sqrt{2}}\\arctan\\left(\\dfrac{\\tan\\dfrac{x}{2}}{\\sqrt{3+2\\sqrt{2}}}\\right)\\right)}{2}$","cooked":"

$\\displaystyle \\int \\dfrac{1+\\cos{x}}{1+\\sin^2{x}}~\\mathrm{d}x$

该不定积分如何计算？看了看 ggb 和 mathematica 的结果，有些复杂，不太像大一能写的

$\"ggb\"$
\n $\"mathematica\"$

感谢门友相助，已解决（但是没找到“解决方案”按钮），下面梳理一下思路

取代换 $t=\\tan\\left(\\dfrac{x}{2}\\right)$ ，得到一个有理函数的不定积分
对于分母中的二次三项式 $x^2+px+q$ ，可先配方得 $\\left(x+\\dfrac{p}{2}\\right)^2-\\dfrac{p^2}{4}+q$ ，再由平方差公式进一步化简，得到两个分式（打完才想起来 $x$ 用过了，懒得改了）
最后查表并化简

最终结果（如果我没算错的话）： $\\dfrac{\\sqrt{2}\\left(\\sqrt{3+2\\sqrt{2}}\\arctan\\left(\\dfrac{\\tan\\dfrac{x}{2}}{\\sqrt{3-2\\sqrt{2}}}\\right)-\\sqrt{3-2\\sqrt{2}}\\arctan\\left(\\dfrac{\\tan\\dfrac{x}{2}}{\\sqrt{3+2\\sqrt{2}}}\\right)\\right)}{2}$

\n","createdAt":"2024-11-18 08:11:06.264043","updatedAt":"2024-11-18 08:59:42.94349","replyToPostNumber":null,"replyCount":0,"likeCount":0,"wordCount":170,"liked":false,"bookmarked":false,"polls":[],"reactions":[],"wiki":false,"version":3,"publicVersion":3,"deletedAt":null,"isAnonymous":false,"user":{"id":5905,"username":"PipaQinse233","name":"PipaQinse233","avatarUrl":"/uploads/image/f7/1b/f71bbda87a32de08df517fe7ee4a14e3f63bfd1b2105d3ddb8ba1f4c2eaa1c29.avif","admin":false,"moderator":false,"signatureRaw":"[right]——發自我的手機[/right]","signatureCooked":"

——發自我的手機

"}},{"id":43344,"postNumber":2,"raw":"纯三角函数找基频这里都化成了 x/2 然后就转换成了有理函数的积分这是有固定套路的","cooked":"

纯三角函数找基频这里都化成了 x/2 然后就转换成了有理函数的积分这是有固定套路的

\n","createdAt":"2024-11-18 08:17:42.821808","updatedAt":"2024-11-18 08:17:42.821808","replyToPostNumber":null,"replyCount":1,"likeCount":0,"wordCount":37,"liked":false,"bookmarked":false,"polls":[],"reactions":[],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":4668,"username":"IO","name":null,"avatarUrl":"/uploads/image/86/f2/86f267c0aaa73835a999b7cc3a7e7a701605574eda9a2c2a56f5ddbbba64d1a9.avif","admin":true,"moderator":false,"signatureRaw":"","signatureCooked":null}},{"id":43346,"postNumber":3,"raw":"就是取 $t = \\tan{\\left(\\dfrac{x}{2}\\right)}$，对吗？","cooked":"

就是取 $t = \\tan{\\left(\\dfrac{x}{2}\\right)}$ ，对吗？

\n","createdAt":"2024-11-18 08:19:18.763302","updatedAt":"2024-11-18 08:19:18.763302","replyToPostNumber":2,"replyCount":0,"likeCount":0,"wordCount":9,"liked":false,"bookmarked":false,"polls":[],"reactions":[],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":5905,"username":"PipaQinse233","name":"PipaQinse233","avatarUrl":"/uploads/image/f7/1b/f71bbda87a32de08df517fe7ee4a14e3f63bfd1b2105d3ddb8ba1f4c2eaa1c29.avif","admin":false,"moderator":false,"signatureRaw":"[right]——發自我的手機[/right]","signatureCooked":"

——發自我的手機

"}},{"id":43347,"postNumber":4,"raw":"https://baike.baidu.com/item/%E6%AD%A3%E5%88%87%E5%8D%8A%E8%A7%92%E5%85%AC%E5%BC%8F/22781837","cooked":"

https://baike.baidu.com/item/%E6%AD%A3%E5%88%87%E5%8D%8A%E8%A7%92%E5%85%AC%E5%BC%8F/22781837

","createdAt":"2024-11-18 08:21:05.284557","updatedAt":"2024-11-18 08:21:05.284557","replyToPostNumber":null,"replyCount":0,"likeCount":1,"wordCount":1,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":1,"users":[{"id":5905,"username":"PipaQinse233","name":"PipaQinse233"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":4668,"username":"IO","name":null,"avatarUrl":"/uploads/image/86/f2/86f267c0aaa73835a999b7cc3a7e7a701605574eda9a2c2a56f5ddbbba64d1a9.avif","admin":true,"moderator":false,"signatureRaw":"","signatureCooked":null}},{"id":43349,"postNumber":5,"raw":"搜索 Weierstrass substitution 有很多教程","cooked":"

搜索 Weierstrass substitution 有很多教程

\n","createdAt":"2024-11-18 08:25:12.749159","updatedAt":"2024-11-18 08:25:12.749159","replyToPostNumber":null,"replyCount":0,"likeCount":1,"wordCount":9,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":1,"users":[{"id":5905,"username":"PipaQinse233","name":"PipaQinse233"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":4668,"username":"IO","name":null,"avatarUrl":"/uploads/image/86/f2/86f267c0aaa73835a999b7cc3a7e7a701605574eda9a2c2a56f5ddbbba64d1a9.avif","admin":true,"moderator":false,"signatureRaw":"","signatureCooked":null}},{"id":43350,"postNumber":6,"raw":"例如 https://math24.net/weierstrass-substitution.html","cooked":"

例如 https://math24.net/weierstrass-substitution.html

\n","createdAt":"2024-11-18 08:25:50.689482","updatedAt":"2024-11-18 08:25:50.689482","replyToPostNumber":null,"replyCount":1,"likeCount":1,"wordCount":3,"liked":false,"bookmarked":false,"polls":[],"reactions":[{"reaction":"❤️","count":1,"users":[{"id":5905,"username":"PipaQinse233","name":"PipaQinse233"}],"hasReacted":false}],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":4668,"username":"IO","name":null,"avatarUrl":"/uploads/image/86/f2/86f267c0aaa73835a999b7cc3a7e7a701605574eda9a2c2a56f5ddbbba64d1a9.avif","admin":true,"moderator":false,"signatureRaw":"","signatureCooked":null}},{"id":43351,"postNumber":7,"raw":"非常感谢","cooked":"

非常感谢

\n","createdAt":"2024-11-18 08:29:17.486384","updatedAt":"2024-11-18 08:29:17.486384","replyToPostNumber":6,"replyCount":0,"likeCount":0,"wordCount":4,"liked":false,"bookmarked":false,"polls":[],"reactions":[],"wiki":false,"version":1,"publicVersion":1,"deletedAt":null,"isAnonymous":false,"user":{"id":5905,"username":"PipaQinse233","name":"PipaQinse233","avatarUrl":"/uploads/image/f7/1b/f71bbda87a32de08df517fe7ee4a14e3f63bfd1b2105d3ddb8ba1f4c2eaa1c29.avif","admin":false,"moderator":false,"signatureRaw":"[right]——發自我的手機[/right]","signatureCooked":"

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