不定积分公式和推导

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不定积分公式

一、记忆部分

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

二、三角函数代换求二次积分

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

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

$\int\frac{dx}{a^2+x^2} = \frac{1}{a}\arctan\frac{x}{a}+C$

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

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

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

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

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

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

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

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

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

三、常用变换

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

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

$1+\cos x = 2\cos^2\frac{x}{2}$ （二倍角公式）
$\int\frac{dx}{1+\cos x} = \int\frac{1-\cos x}{1-\cos^2x}dx = \int\frac{1-\cos x}{\sin^2x}dx = \frac{1-\cos x}{\sin x}+C$
待更新

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不定积分公式和推导