不定積分公式

一、記憶部分

$\int \tan xdx=?\ln |\cos x|+C$

$\int \cot xdx=\ln |\sin x|+C$

$\int \sec xdx=\int \frac{1}{\cos x}dx=\ln |\frac{1+\sin x}{\cos x}|+C$

$\int \csc xdx=\int \frac{1}{\sin x}dx=\ln |\frac{1?\cos x}{\sin x}|+C$

$\int se{c}^{2}xdx=\int \frac{1}{{\cos }^{2}x}dx=\tan x+C$

$\int cs{c}^{2}dx=\int \frac{1}{{\sin }^{2}x}dx=?\cot x+C$

二、三角函數(shù)代換求二次積分

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

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

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

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

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

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

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

$$\begin{gathered} \int \frac{dx}{\sqrt{x^2?a^2}}\underline{\underline{x=a\sec t}}\int \frac{da\sec t}{\sqrt{a^2{\sec }^{2}t?a^2}}=\int \frac{a\sec t\tan tdt}{a\tan t}=\int \sec tdt=\ln |\frac{1+\sin t}{\cos t}|+C \\ \because x=a\sec t(畫三角形） \\ \therefore \sec t=\frac{x}{a},\tan t=\frac{\sqrt{x^2?a^2}}{a} \\ \therefore \ln |\frac{1+\sin t}{\cos t}|+C=\ln |\sec t+\tan t|+C=\ln |x+\sqrt{x^2?a^2}|?\ln |a|+C=\ln (x+\sqrt{x^2?a^2})+C \end{gathered}$$

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

$$\begin{gathered} \int \frac{dx}{\sqrt{x^2+a^2}}\underline{\underline{x=a\tan t}}\int \frac{da\tan t}{\sqrt{a^2{\tan }^{2}t+a^2}}=\int \frac{a{\sec }^{2}tdt}{a\sec t}=\int \sec tdt=\ln |\frac{1+\sin t}{\cos t}|+C \\ \because x=a\tan t(畫三角形） \\ \therefore \tan t=\frac{x}{a},\sec t=\frac{\sqrt{x^2+a^2}}{a} \\ \therefore \ln |\frac{1+\sin t}{\cos t}|+C=\ln |\sec t+\tan t|+C=\ln |x+\sqrt{x^2+a^2}|?\ln |a|+C=\ln (x+\sqrt{x^2+a^2})+C \end{gathered}$$

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

$$\begin{gathered} \int \sqrt{a^2?x^2}dx=x\sqrt{a^2?x^2}?\int xd\sqrt{a^2?x^2}=x\sqrt{a^2?x^2}+\int \frac{x^2}{\sqrt{a^2?x^2}}dx=x\sqrt{a^2?x^2}+\int \frac{x^2?a^2+a^2}{\sqrt{a^2?x^2}}dx \\ =x\sqrt{a^2?x^2}?\int \sqrt{a^2?x^2}dx+\int \frac{a^2}{\sqrt{a^2?x^2}}dx \\ \therefore 2\int \sqrt{a^2?x^2}dx=x\sqrt{a^2?x^2}+a^2\arcsin \frac{x}{a}+C \\ \therefore \int \sqrt{a^2?x^2}dx=\frac{a^2}{2}\arcsin \frac{x}{a}+\frac{1}{2}x\sqrt{a^2?x^2}+C \end{gathered}$$

三、常用變換

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

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

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

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

待更新

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