不定积分公式

一、记忆部分

  1. $\int{\tan x}dx = -\ln |\cos x|+C$
  2. $\int \cot xdx = \ln |\sin x|+C$
  3. $\int \sec x dx =\int \frac{1}{\cos x}dx= \ln | \frac{1+\sin x}{\cos x}|+C$
  4. $\int \csc xdx = \int \frac{1}{\sin x}dx = \ln |\frac{1-\cos x}{ \sin x}|+C$
  5. $\int sec^2xdx = \int \frac{1}{\cos ^2x}dx = \tan x +C$
  6. $\int csc^2dx = \int \frac{1}{\sin^2x}dx = -\cot x+C$

二、三角函数代换求二次积分

  1. $\int \frac{dx}{\sqrt{a2-x2}} = \arcsin \frac{x}{a}+C$

$$
\int \frac{dx}{\sqrt{a2-x2}},\underline{\underline{x =a\sin t}},\int\frac{d,a\sin t}{\sqrt{a2-a2\sin^2t}} = \int 1,dt = t+C ,,,,\underline{\underline{t = \arcsin \frac{x}{a}}},,,,\arcsin x+C
$$

  1. $\int\frac{dx}{a2+x2} = \frac{1}{a}\arctan\frac{x}{a}+C$

$$
\int\frac{dx}{a2+x2},\underline{\underline{x =a\tan t}},\int\frac{d,a\tan t}{a2+a2\tan^2t} = \int\frac{a\sec2tdt}{a2\sec^2t} = \frac {t}{a}+C ,,,,\underline{\underline{t = \arctan \frac{x}{a}}},,,, \frac{1}{a}\arctan \frac{x}{a}+C
$$

  1. $\int \frac{dx}{x2-a2} = \frac{1}{2a}\ln|\frac{x-a}{x+a}|+C$

$$
\int \frac{dx}{x2-a2} = \int \frac{dx}{(x-a)(x+a)} = \frac{1}{2a}\int (\frac{1}{x-a}-\frac{1}{x+a})da = \frac{1}{2a}[\ln|x-a|-ln|x+a|]+C = \frac{1}{2a}\ln|\frac{x-a}{x+a}|+C
$$

  1. $\int \frac{dx}{\sqrt{x2-a2}} = \ln(x+\sqrt{x2-a2})+C$

$$
\int \frac{dx}{\sqrt{x2-a2}} ,,,,\underline{\underline{x = a\sec t}},,,,\int\frac{d,a\sec t}{\sqrt{a2\sec2t-a^2}} = \int\frac{a\sec t \tan t,dt}{a\tan t} = \int \sec t,dt =\ln |\frac{1+\sin t}{ \cos t}|+C \
\because x = a\sec t(画三角形）\
\therefore \sec t = \frac{x}{a},,,,,,,,,,\tan t = \frac{\sqrt{x2-a2}}{a}\
\therefore\ln |\frac{1+\sin t}{ \cos t}|+C = \ln|\sec t+\tan t|+C = \ln|x+\sqrt{x2-a2}|-\ln|a|+C = \ln(x+\sqrt{x2-a2})+C
$$

  1. $\int \frac{dx}{\sqrt{x2+a2}} = \ln(x+\sqrt{x2+a2})+C$

$$
\int \frac{dx}{\sqrt{x2+a2}} ,,,,\underline{\underline{x = a\tan t}},,,,\int\frac{d,a\tan t}{\sqrt{a2\tan2t+a^2}} = \int\frac{a\sec^2t ,dt}{a\sec t} = \int \sec t,dt =\ln |\frac{1+\sin t}{ \cos t}|+C \
\because x = a\tan t(画三角形）\
\therefore \tan t = \frac{x}{a},,,,,,,,,,\sec t = \frac{\sqrt{x2+a2}}{a}\
\therefore\ln |\frac{1+\sin t}{ \cos t}|+C = \ln|\sec t+\tan t|+C = \ln|x+\sqrt{x2+a2}|-\ln|a|+C = \ln(x+\sqrt{x2+a2})+C
$$

  1. $\int\sqrt{a2-x2}dx = \frac{a2}{2}\arcsin{\frac{x}{a}}+\frac{1}{2}x\sqrt{a2-x^2}+C$

$$
\int\sqrt{a2-x2}dx = x\sqrt{a2-x2}-\int xd\sqrt{a2-x2} = x\sqrt{a2-x2} + \int \frac{x2}{\sqrt{a2-x^2}}dx = x\sqrt{a2-x2} + \int \frac{x2-a2+a2}{\sqrt{a2-x^2}}dx\
= x\sqrt{a2-x2} - \int \sqrt{a2-x2}dx+\int \frac{a2}{\sqrt{a2-x^2}}dx\
\therefore 2\int\sqrt{a2-x2}dx = x\sqrt{a2-x2}+a^2\arcsin \frac{x}{a}+C\
\therefore \int\sqrt{a2-x2}dx = \frac{a2}{2}\arcsin{\frac{x}{a}}+\frac{1}{2}x\sqrt{a2-x^2}+C
$$

三、常用变换

  1. $\sin x = \frac{2\tan \frac{x}{2}}{1+\tan^2\frac{x}{2}}$

$$
\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2} = \frac{2\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}}{\frac{1}{\cos^2\frac{x}{2}}} = \frac{2\tan \frac{x}{2}}{1+\tan^2\frac{x}{2}}
$$

  1. $1+\cos x = 2\cos^2\frac{x}{2}$（二倍角公式）

  2. $\int\frac{dx}{1+\cos x} = \int\frac{1-\cos x}{1-\cos^2x}dx = \int\frac{1-\cos x}{\sin^2x}dx = \frac{1-\cos x}{\sin x}+C$

  3. 待更新