計算不定積分：$\int \frac{dx}{(2+\cos x)\sin x} $

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湊一下微分（上下乘個 \(\sin x\)，這樣的話上面就有 \(\cos x\) 了，下面用三角恆等變換都弄成 \(\cos x\)）：

\[\begin{aligned} \int \frac{dx}{(2+\cos x)\sin x} = & \int \frac{\sin x dx}{(2+\cos x)\sin^2x} \\ = & -\int \frac{d(\cos x)}{(2+\cos x)(1-\cos ^2x)} \\ = & -\int \left(\frac{-\frac{1}{3}}{2+\cos x} + \frac{\frac{1}{2}}{1+\cos x} + \frac{\frac{1}{6}}{1-\cos x} \right)d (\cos x) \\ = & \frac{1}{3}\ln|2+\cos x|-\frac{1}{2} \ln |1+\cos x|+\frac{1}{6}\ln|1-\cos x| + C\\ = & \frac{1}{6}\ln \left| \frac{(1-\cos x)(2+\cos x)^2}{(1+\cos x)^3} \right| + C \end{aligned} \]

當然也可以直接簡單粗暴的萬能代換：

\[\begin{aligned} \int \frac{dx}{(2+\cos x) \sin x} \xlongequal{u=\tan \frac{x}{2}} & \int \frac{\frac{2du}{1+u^2}}{(2+\frac{1-u^2}{1+u^2})\frac{2u}{1+u^2}} \\ =&\int \frac{1+u^2}{3+u^2} \cdot \frac{du}{u} \\ =& \int \left(\frac{\frac{1}{3}}{u} + \frac{\frac{2}{3}u}{3+u^2} \right)du \\ =&\frac{1}{3}\ln|u|+\frac{1}{3}\ln |3+u^2| + C \\ =&\frac{1}{3} \ln \left|\left(3+\tan^2\frac{x}{2}\right)\tan \frac{x}{2}\right|+C \end{aligned} \]

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這時候會發現結果部分有些不太對！

\[\frac{1}{6}\ln \left| \frac{(1-\cos x)(2+\cos x)^2}{(1+\cos x)^3} \right|=\frac{1}{3} \ln \left|\sqrt{\frac{(2\sin^2\frac{x}{2})(3-2\sin^2\frac{x}{2})^2}{(2\cos^2\frac{x}{2})^3}}\right|=\frac{1}{3}\ln\left| \frac{1}{2} \cdot \frac{\tan\frac{x}{2}(3-2\sin^2\frac{x}{2})}{\cos^2\frac{x}{2}} \right| \]\[\frac{1}{3} \ln \left|\left(3+\tan^2\frac{x}{2}\right)\tan \frac{x}{2}\right|=\frac{1}{3}\ln \left| \frac{3\cos^2 \frac{x}{2}+\sin^2\frac{x}{2}}{\cos^2 \frac{x}{2}} \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} \right|=\frac{1}{3}\ln \left| \frac{3-2\sin^2\frac{x}{2}}{\cos^2 \frac{x}{2}} \tan \frac{x}{2} \right| \]

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然而真的不太對嘛？並不：\(\frac{1}{3}\ln |\frac{1}{2}\cdot t|=\frac{1}{3}\ln |t|+\frac{1}{3}\ln\frac{1}{2}=\frac{1}{3}\ln|t|+C'\)

所以實際上沒算錯（這個常數問題真是坑……）