Find and fix gaps in Volume 1 of Calculus.

常用积分公式(积分表)

【Know Them Thoroughly】

$\int a^x \, dx = \frac{a^x}{\ln a} + C \quad (a > 0, a ≠ 1)$

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

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

$\int \sec^2 x \, dx = \tan x + C$

$\int \csc^2 x \, dx = -\cot x + C$

$\int \sec x \tan x \, dx = \sec x + C$

$\int \csc x \cot x \, dx = -\csc x + C$

$\int \sec x \, dx = \ln|\sec x + \tan x| + C$

$\int \csc x \, dx = \ln|\csc x - \cot x| + C = -\ln|\csc x + \cot x| + C$ The two forms of expression can be converted into each other.

【考前背一背组之三角函数】关键在于:知道左边的形式有公式!做题往左边凑!

$\int \tan x \, dx = -\ln|\cos x| + C$

$\int \cot x \, dx = \ln|\sin x| + C$

$\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \arctan\frac{x}{a} + C \quad (a > 0)$

$\int \frac{dx}{\sqrt{x^2 + a^2}} = \ln\left|x + \sqrt{x^2 + a^2}\right| + C \quad (a > 0)$

$\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\frac{x}{a} + C \quad (a > 0)$

$\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \ln\left|\frac{a + x}{a - x}\right| + C \quad (a > 0)$ (不用三角函数换元也可以做出来,直接分母拆分就可以了。)