积分技巧

$\displaystyle\int \sec xdx=\ln|\sec x+\tan x|+C$

证明：

$$\begin{align*} \int \sec xdx=\int\frac{1}{\cos x}dx &=\int\frac{\sec x+\tan x}{\cos x(\sec x+\tan x)}dx\\&=\int\frac{1}{\sec x+\tan x}d(\sec x+\tan x)\\&=\ln|\sec x+\tan x|+C \end{align*}$$

或：

$$\begin{align*} \int \sec xdx=\int\frac{\cos x}{\cos^2 x}dx &=\int\frac{1}{1-\sin^2 x}d\sin x\\&=\frac{1}{2}\int\left(\frac{1}{1+\sin x}+\frac{1}{1-\sin x}\right)d\sin x\\&=\frac{1}{2}\int\left(\frac{1}{1+\sin x}\right)d(1+\sin x)-\frac{1}{2}\int\left(\frac{1}{1-\sin x}\right)d(1-\sin x)\\&=\frac{1}{2}\ln\left|\frac{1+\sin x}{1-\sin x}\right|+C \end{align*}$$

$\displaystyle\int \csc xdx=\ln|\csc x-\cot x|+C$

证明：

$$\begin{align*}\int \csc xdx=\int\frac{1}{\sin x}&=\int\frac{\csc x-\cot x}{\sin x(\csc x-\cot x)}\\&=\int\frac{1}{\csc x-\cot x}d(\csc x-\cot x)\\&=\ln|\csc x-\cot x|+C \end{align*}$$

或：

$$\begin{align*}\displaystyle\int \csc xdx=\int\frac{\sin x}{\sin ^2x}dx&=\int\frac{1}{1-\cos^2}d\sin x\\&=\frac{1}{2}\left(\int\frac{1}{1-\cos x}d(1-\cos x)-\int\frac{1}{1+\cos x}d(1+\cos x)\right)\\&=\frac{1}{2}\ln\left|\frac{1-\cos x}{1+\cos x}\right| \end{align*}$$

$\displaystyle\int\frac{dx}{\sqrt{x^2+a^2}}=\ln(x+\sqrt{x^2+a^2})+C$

证明：

由恒等式$1+\tan^2x=\sec^2 x$,令$x=a\tan t$,则$dx=a\sec^2tdt$

则有

$$\begin{align*}\int\frac{dx}{\sqrt{x^2+a^2}}=\int\frac{a\sec^2tdt}{\sqrt{a^2(1+\tan^2t)}}&=\int\sec tdt\\&=\ln|\sec t+\tan t|+C\\&=\ln|\frac{\sqrt{a^2+x^2}+x}{a}|+C\\&=\ln(x+\sqrt{x^2+a^2})+C^{'} \end{align*}$$

$$\int \frac{\cos t}{\sin^2 t} dt ==-\int \frac{1}{\sin t} dt =-\int \csc t dt=-\ln(\csc t-\cot t)$$