三角函数分式积分

$$\int \frac{dx}{\sin x}$$

$$\begin{array}{rl}& =\int \frac{-d\cos x}{1-{\cos }^{2}x}\\ & =-\int \frac{1}{2}(\frac{1}{1+\cos x}+\frac{1}{1-\cos x})d\cos x\\ & =-\frac{1}{2}\ln \left|\frac{1+\cos x}{1-\cos x}\right|+C\\ & =-\ln \left|\frac{1+\cos x}{\sin x}\right|+C\end{array}$$

$$\int \frac{dx}{{\sin }^{2}x}$$

$$\begin{array}{rl}& =\int \frac{d\tan x}{{\tan }^{2}x}\\ & =-\frac{1}{\tan x}+C\end{array}$$

$$\int \frac{dx}{{\sin }^{3}x}$$

有理分式裂项是可以做的

$$\begin{array}{rl}& =\int \frac{-d\cos x}{{\sin }^{4}x}\\ & =-\int \frac{d\cos x}{(1-{\cos }^{2}x{)}^2}\\ & =-\int (\frac{A}{(t-1{)}^2}+\frac{B}{t-1}+\frac{C}{(t+1{)}^2}+\frac{D}{t+1})dx\\ & =-\int (\frac{1/4}{(t-1{)}^2}+\frac{-1/4}{t-1}+\frac{1/4}{(t+1{)}^2}+\frac{1/4}{t+1})dx\\ & =-\frac{1}{4}(-\frac{1}{t-1}-\ln |t-1|-\frac{1}{t+1}+\ln |t+1\left|\right)+C\\ & =-\frac{1}{4}(-\frac{2t}{t^2-1}+\ln |\frac{t+1}{t-1}|)+C\\ & =-\frac{1}{4}(\frac{2\cos x}{{\sin }^{2}x}+\ln |\frac{\cos x+1}{\cos x-1}|)+C\end{array}$$

或者可以采用分部积分

$$\begin{array}{rl}& =\int \frac{-1}{\sin x}d\frac{\cos x}{\sin x}\\ & =-\frac{\cos x}{{\sin }^{2}x}+C-\int \frac{\cos x}{\sin x}d\frac{-1}{\sin x}\\ & =-\frac{\cos x}{{\sin }^{2}x}+C-\int \frac{{\cos }^{2}x}{{\sin }^{3}x}dx\\ & =-\frac{\cos x}{{\sin }^{2}x}+C-\int \frac{1-{\sin }^{2}x}{{\sin }^{3}x}dx\\ & =-\frac{\cos x}{{\sin }^{2}x}+C-\int \frac{1}{{\sin }^{3}x}dx+\int \frac{1}{\sin x}dx\\ & =-\frac{\cos x}{2{\sin }^{2}x}-\frac{1}{2}\ln \left|\frac{1+\cos x}{\sin x}\right|+C\end{array}$$

$$\int \frac{1}{{\sin }^{4}x}dx$$

$$\begin{array}{rl}& =\int \frac{-1}{{\sin }^{2}x}d\frac{\cos x}{\sin x}\\ & =-\frac{\cos x}{{\sin }^{3}x}+C-\int \frac{\cos x}{\sin x}d\frac{-1}{{\sin }^{2}x}\\ & =-\frac{\cos x}{{\sin }^{3}x}+C-\int \frac{2{\cos }^{2}x}{{\sin }^{4}x}dx\\ & =-\frac{\cos x}{{\sin }^{3}x}+C-2\int \frac{1-{\sin }^{2}x}{{\sin }^{4}x}dx\\ & =-\frac{\cos x}{{\sin }^{3}x}+C-2\int \frac{1}{{\sin }^{4}x}dx+\int \frac{2}{{\sin }^{2}x}dx\\ & =-\frac{\cos x}{3{\sin }^{3}x}-\frac{2}{3\tan x}+C\end{array}$$

可以看到更高次项同理

$$\int {\tan }^{n}xdx$$

$$\begin{array}{rl}& =\int \frac{{\sin }^{n-2}x(1-{\cos }^{2}x)}{{\cos }^{n}x}dx\\ & =\int {\tan }^{n-2}xd\tan x+\int {\tan }^{n-2}xdx\\ & =\frac{{\tan }^{n-1}x}{n-1}+\int {\tan }^{n-2}xdx+C\end{array}$$

形成递归，边界分别为

1. $\int \tan xdx=-\ln \left|\cos x\right|+C$
2. $\int dx=x+C$