1/sinx+cosx的不定积分 1/sinx+cosx 的不定积分是什么，如何推导的
1/(sinx+cosx)\u7684\u4e0d\u5b9a\u79ef\u5206\u600e\u4e48\u6c42\uff1f\u4ee4u = tan(x / 2),dx = 2du / (1+u²)
sinx = 2u / (1+u²),cosx = (1 - u²) / (1 + u²)
\u222b dx / (sinx + cosx)
= \u222b 2 / \u3010(1 + u²) * [2u / (1+u²) + (1 - u²) / (1 + u²)]\u3011 du
= 2\u222b du / (-u² + 2u + 1)
= 2\u222b du / [2 - (u - 1)²]
= 2\u222b dy / (2 - y²),y=u - 1
= (1 / 2\u221a2)ln|(y + \u221a2) / (y - \u221a2)| + C
= (1 / 2\u221a2)ln|(u - 1 + \u221a2) / (y - 1 - \u221a2)| + C
= (1 / 2\u221a2)ln|[tan(x / 2) - 1 + \u221a2] / [tan(x / 2) - 1 - \u221a2)| + C
= \u221a2arctanh\u3010[tan(x / 2) - 1] / \u221a2\u3011+ C
\u6269\u5c55\u8d44\u6599\u7b2c\u4e8c\u7c7b\u6362\u5143\u6cd5\u7ecf\u5e38\u7528\u4e8e\u6d88\u53bb\u88ab\u79ef\u51fd\u6570\u4e2d\u7684\u6839\u5f0f\u3002\u5f53\u88ab\u79ef\u51fd\u6570\u662f\u6b21\u6570\u5f88\u9ad8\u7684\u4e8c\u9879\u5f0f\u7684\u65f6\u5019\uff0c\u4e3a\u4e86\u907f\u514d\u7e41\u7410\u7684\u5c55\u5f00\u5f0f\uff0c\u6709\u65f6\u4e5f\u53ef\u4ee5\u4f7f\u7528\u7b2c\u4e8c\u7c7b\u6362\u5143\u6cd5\u6c42\u89e3\u3002\u5e38\u7528\u7684\u6362\u5143\u624b\u6bb5\u6709\u4e24\u79cd\uff1a
1\u3001 \u6839\u5f0f\u4ee3\u6362\u6cd5\uff0c
2\u3001 \u4e09\u89d2\u4ee3\u6362\u6cd5\u3002
\u5728\u5b9e\u9645\u5e94\u7528\u4e2d\uff0c\u4ee3\u6362\u6cd5\u6700\u5e38\u89c1\u7684\u662f\u94fe\u5f0f\u6cd5\u5219\uff0c\u800c\u5f80\u5f80\u7528\u6b64\u4ee3\u66ff\u524d\u9762\u6240\u8bf4\u7684\u6362\u5143\u3002
\u94fe\u5f0f\u6cd5\u5219\u662f\u4e00\u79cd\u6700\u6709\u6548\u7684\u5fae\u5206\u65b9\u6cd5\uff0c\u81ea\u7136\u4e5f\u662f\u6700\u6709\u6548\u7684\u79ef\u5206\u65b9\u6cd5\u3002
\u4ee4u = tan(x / 2),dx = 2du / (1+u²)
sinx = 2u / (1+u²),cosx = (1 - u²) / (1 + u²)
\u222b dx / (sinx + cosx)
= \u222b 2 / \u3010(1 + u²) * [2u / (1+u²) + (1 - u²) / (1 + u²)]\u3011 du
= 2\u222b du / (-u² + 2u + 1)
= 2\u222b du / [2 - (u - 1)²]
= 2\u222b dy / (2 - y²),y=u - 1
= (1 / 2\u221a2)ln|(y + \u221a2) / (y - \u221a2)| + C
= (1 / 2\u221a2)ln|(u - 1 + \u221a2) / (y - 1 - \u221a2)| + C
= (1 / 2\u221a2)ln|[tan(x / 2) - 1 + \u221a2] / [tan(x / 2) - 1 - \u221a2)| + C
= \u221a2arctanh\u3010[tan(x / 2) - 1] / \u221a2\u3011+ C
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