证明
\[
\int\sec xdx=\int\frac{1}{\cos x}dx=\ln |\sec x+\tan x|+C.
\]
\[
\int\csc xdx=\int\frac{1}{\sin x}dx=\frac{1}{2}\ln\biggl|\frac{1-\cos x}{1+\cos x}\biggr|+C=\ln\biggl|\tan\frac{x}{2}\biggr|+C.
\]
并求 $\int \sec^2 xdx$, $\int \sec^3 xdx$
[总结]
\[
\begin{aligned}
\int\sec xdx&=\ln|\sec x+\tan x|+C\\
\int\sec^2 xdx&=\tan x+C\\
\int\sec^3 xdx&=\frac{1}{2}\ln|\sec x+\tan x|+\frac{1}{2}\sec x\cdot\tan x+C
\end{aligned}
\]
1
(法一)
\[
\begin{split}
\int\csc xdx&=\int\frac{1}{\sin x}dx=\int\frac{\sin x}{\sin^2 x}dx=-\int\frac{d\cos x}{1-\cos^2 x}\\
&\stackrel{t=\cos x}{=}\int\frac{dt}{t^2-1}=\frac{1}{2}\int(\frac{1}{t-1}-\frac{1}{t+1})dt\\
&=\frac{1}{2}\ln\biggl|\frac{t-1}{t+1}\biggr|+C\\
&=\frac{1}{2}\ln\biggl|\frac{1-\cos x}{1+\cos x}\biggr|+C.
\end{split}
\]
(法二)
\[
\begin{split}
\int\csc xdx&=\int\frac{1}{\sin x}dx=\int\frac{dx}{2\sin\frac{x}{2}\cos\frac{x}{2}}\\
&=\frac{1}{2}\int\frac{dx}{\tan\frac{x}{2}\cos^2\frac{x}{2}}=\frac{1}{2}\int\frac{1}{\tan\frac{x}{2}}\cdot\sec^2\frac{x}{2}dx\\
&=\int\frac{1}{\tan\frac{x}{2}}d\tan\frac{x}{2}\\
&\stackrel{t=\tan\frac{x}{2}}{=}\int\frac{1}{t}dt=\ln|t|+C\\
&=\ln\biggl|\tan\frac{x}{2}\biggr|+C.
\end{split}
\]
答案虽然从形式上看不一致, 但其实是相同的. 事实上,
\[
\begin{split}
\frac{1}{2}\ln\biggl|\frac{1-\cos x}{1+\cos x}\biggr|&=\frac{1}{2}\ln\biggl|\frac{(1-\cos x)^2}{1-\cos^2 x}\biggr|\\
&=\ln\biggl|\frac{1-\cos x}{\sin x}\biggr|\\
&=\ln\biggl|\frac{2\sin^2\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}}\biggr|\\
&=\ln\biggl|\tan\frac{x}{2}\biggr|.
\end{split}
\]
2
\[
\int\sec^2 xdx=\tan x+C
\]
\[
\begin{split}
\int\sec^3 xdx&=\int\sec xd\tan x\\
&=\sec x\tan x-\int\tan xd\sec x\\
&=\sec x\tan x-\int\tan^2 x\sec xdx\\
&=\sec x\tan x-\int(\sec^2 x -1)\sec xdx\\
&=\sec x\tan x-\int\sec^3 xdx+\int\sec xdx\\
\end{split}
\]
因此
\[
\int\sec^3 xdx=\frac{1}{2}\sec x\tan x+\frac{1}{2}\int\sec xdx
\]
而 $\int\sec xdx=\ln|\sec x+\tan x|+C$, 故
\[
\int\sec^3 xdx=\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln|\sec x+\tan x|+C
\]
3
上面求 $\int\sec^3 xdx$ 用了分部积分. 如果尚未讲分部积分, 则可以使用换元法.
\[
\int\sec^3 xdx=\int\frac{1}{\cos^3 x}dx=\int\frac{\cos x}{\cos^4 x}dx=\int\frac{d\sin x}{(1-\sin^2 x)^2}
\]
令 $t=\sin x$, 则上式等于
\[
\begin{split}
\int\frac{dt}{(1-t^2)^2}&=\int\biggl[\frac{1}{2}\bigl(\frac{1}{1-t}+\frac{1}{1+t}\bigr)\biggr]^2dt\\
&=\frac{1}{4}\int\Bigl(\frac{1}{(1-t)^2}+\frac{2}{1-t^2}+\frac{1}{(1+t)^2}\Bigr)dt\\
&=\frac{1}{4}\biggl[\int\frac{1}{(1-t)^2}dt+\int\frac{2}{1-t^2}dt+\int\frac{1}{(1+t)^2}dt\biggr]\\
&=\frac{1}{4}\biggl[\int\frac{1}{(t-1)^2}d(t-1)+\int(\frac{1}{1-t}+\frac{1}{1+t})dt+\int\frac{1}{(t+1)^2}d(t+1)\biggr]\\
&=\frac{1}{4}\biggl[-\frac{1}{t-1}+\ln|t+1|-\ln|t-1|+(-\frac{1}{t+1})\biggr]+C\\
&=\frac{1}{4}\biggl[\ln\biggl|\frac{t+1}{t-1}\biggr|-\frac{2t}{t^2-1}\biggr]+C\\
&=\frac{1}{4}\biggl[\ln\biggl|\frac{\sin x+1}{\sin x-1}\biggr|-\frac{2\sin x}{\sin^2 x-1}\biggr]+C\\
&=\frac{1}{4}\biggl[\ln\biggl|\frac{1+\sin x}{1-\sin x}\biggr|+\frac{2\sin x}{\cos^2 x}\biggr]+C\\
&=\frac{1}{4}\biggl[\ln\biggl|\frac{(1+\sin x)^2}{1-\sin^2 x}\biggr|+2\frac{\sin x}{\cos x}\cdot\frac{1}{\cos x}\biggr]+C\\
&=\frac{1}{4}\ln\Bigl(\frac{1+\sin x}{\cos x}\Bigr)^2+\frac{1}{2}\sec x\cdot\tan x+C\\
&=\frac{1}{2}\ln\biggl|\frac{1+\sin x}{\cos x}\biggr|+\frac{1}{2}\sec x\cdot\tan x+C\\
&=\frac{1}{2}\ln|\sec x+\tan x|+\frac{1}{2}\sec x\cdot\tan x+C\\
\end{split}
\]