常用的积分技巧。下面的 \(\text{d}x\) 都懒得打了，直接用 \(dx\) 代替。

Basics

Rules

\[\begin{aligned} \int f(x)\pm g(x) dx&=\int f(x)dx\pm \int g(x)dx\\ \int f(x)g'(x)dx&=f(x)g(x)-\int g(x)f'(x)dx\\ \int f'(g(x))g'(x)dx&=\int f'(g(x))dg(x)=f(g(x))+C \end{aligned} \]

Exp & Log & Pow

\[\begin{aligned} \int\frac{dx}x&=\ln|x|+C\\ \int\exp(x)&=\exp(x)+C\\ \int x^adx&=\frac{x^{a+1}}{a+1}+C \end{aligned} \]

不用解释。

\[\begin{aligned} \int\ln(x)dx&=\int\ln(x)\cdot 1dx\\ &=x\ln x-\int 1dx\\ &=x\ln x-x+C \end{aligned} \]

Trig

\[\begin{aligned} \int\sin(x)dx&=-\cos(x)dx+C\\ \int\cos(x)dx&=\sin(x)dx+C\\ \end{aligned} \]

不用解释。

\[\begin{aligned} \int\tan(x)dx&=\int\frac{\sin x}{\cos x}dx\\ &=\int\frac1{\cos x}\sin xdx\\ &=-\int\frac1{\cos x}d(\cos x)\\ &=-\ln|\cos(x)|+C \end{aligned} \]

同理

\[\int\cot(x)dx=\ln|\sin(x)|+C \]

对于 \(\sec\) 函数，我们想要寻找 \(f(x)\) 使得 \(\frac{f'(x)}{f(x)}=\sec x\)。经过一些尝试可以得到 \(f(x)=\tan(x)+\sec(x)\)。于是

\[\int\sec(x)dx=\ln|\tan(x)+\sec(x)|+C \]

也可以通过这样：

\[\begin{aligned} \int\sec(x)dx&=\int\frac1{\cos(x)}dx\\ &=\int\frac{\cos(x)}{1-\sin^2(x)}dx\\ &=\int\frac1{1-\sin^2(x)}d(\sin x)\\ &=\frac12(\int\frac1{1+\sin(x)}d(\sin x)-\int\frac1{1-\sin(x)}d(\sin x))\\ &=\frac12\ln\frac{1+\sin(x)}{1-\sin(x)}+C\\ &=\ln|\frac{1+\sin(x)}{\cos(x)}|+C\\ \end{aligned} \]

同理

\[\int\csc(x)dx=-\ln|\cot(x)+\csc(x)|+C \]

Advanced

\[\begin{aligned} \int\sin^3(x)dx&=\int\sin(x)(1-\cos^2(x))dx\\ &=\int\sin(x)dx-\int\sin(x)\cos^2(x)dx\\ &=-\cos(x)+\int\cos^2(x)d(\cos x)\\ &=\frac{\cos^3(x)}3-\cos(x)+C\\\\ \int\sin^4(x)dx&=\frac14\int(1-\cos(2x))^2dx\\ &=\frac18\int(1-\cos(2x))^2d(2x)\\ &=\frac18\int1-2\cos(2x)+\cos^2(2x)d(2x)\\ &=\frac14x-\frac14\sin(2x)+\frac1{16}\int(1+\cos(4x))d(2x)\\ &=\frac38x-\frac14\sin(2x)+\frac1{32}\sin(4x)+C\\\\ \int\sec^3(x)dx&=\int\sec^2(x)\sec(x)dx\\ &=\sec(x)\tan(x)-\int\tan(x)\sec(x)\tan(x)dx\\ &=\sec(x)\tan(x)-\int(\sec^2(x)-1)\sec(x)dx\\ &=\sec(x)\tan(x)\int\sec^3(x)dx+\int\sec(x)dx\\ &=\frac12\sec(x)\tan(x)+\frac12\int\sec(x)dx\\ &=\frac12\sec(x)\tan(x)+\frac12\ln|\tan(x)+\sec(x)|+C\\\\ \int\frac{x+\sin x}{1+\cos x}dx&=\int\frac{x+2\sin\frac x2\cos\frac x2}{2\cos^2\frac x2}dx\\ &=\int\tan\frac x2dx+\int\frac{\frac x2}{\cos^2\frac x2}dx\\ &=2\int\tan\frac x2d(\frac x2)+2(\frac x2\tan\frac x2-\int\tan\frac x2d(\frac x2))\\ &=x\tan\frac x2+C\\ &=\frac{x\sin x}{1+\cos x}+C\\\\ \int\frac{x+\cos x}{1+\sin x}dx&=\int\frac{\cos x}{1+\sin x}dx+\int\frac x{1+\sin x}dx\\ &=\ln(1+\sin x)+\int\frac x{1+\sin x}dx\\ &=\ln(1+\sin x)+\int\frac{x(1-\sin x)}{\cos^2 x}dx\\ &=\ln(1+\sin x)+\int\frac x{\cos^2 x}dx-\int\frac{x\sin x}{\cos^2 x}dx\\ &=\ln(1+\sin x)+x\tan x-\int\tan xdx-\int x\tan x\sec xdx\\ &=\ln(1+\sin x)+x\tan x+\ln|\cos x|-x\sec x+\int\sec xdx\\ &=\ln(1+\sin x)+x\tan x+\ln|\cos x|-x\sec x+\ln|\frac{1+\sin x}{\cos x}|+C\\ &=2\ln(1+\sin x)+x\tan x-x\sec x+C\\ &=2\ln(1+\sin x)-\frac{x\cos x}{1+\sin x}+C\\\\ \int\frac{(\cos x-1)^2}{1+\sin x}dx&=\int\frac{\cos^2x}{1+\sin x}dx-2\int\frac{\cos x}{1+\sin x}dx+\int\frac1{1+\sin x}dx\\ &=\int(1-\sin x)dx-2\ln(1+\sin x)+\int\frac{1-\sin x}{\cos^2 x}dx\\ &=x+\cos x-2\ln(1+\sin x)+\tan x-\sec x+C \end{aligned} \]

总结：看到分母出现 \(1\pm\sin x\) 或者 \(1\pm\cos x\) 考虑使用 Pythagorean 定理。当一个积分可以被表达为 \(x\) 乘上 \(\sec^2x,\csc^2x,\sec x\tan x,\csc x\cot x\) 之一，则考虑分部积分。