换元积分法

Integration By Substitution

说明

第一类换元法：要求熟练掌握三角公式
第二类换元法：主要侧重构造三角形，进行三角函数换元

核心是通过凑微分来换元：

\begin{array}{r} \int f [φ (x)] φ^{'} (x) d x = {[\int f (u) d u]}_{u = φ (x)} \end{array}

\begin{array}{r} \int \frac{1}{a^{2} + x^{2}} d x = \frac{1}{a} \int \frac{1}{1 + (\frac{x}{a})^{2}} d \frac{x}{a} = \frac{1}{a} \arctan \frac{x}{a} + C \end{array}

\begin{array}{r} \int \frac{d x}{\sqrt{a^{2} - x^{2}}} = \int \frac{d \frac{x}{a}}{\sqrt{1 - (\frac{x}{a})^{2}}} = \arcsin \frac{x}{a} + C \end{array}

\begin{array}{r} \int \frac{1}{x^{2} - a^{2}} d x = \int \frac{1}{2 a} (\frac{1}{x - a} - \frac{1}{x + a}) d x = \frac{1}{2 a} \ln | \frac{x - a}{x + a} | + C \end{array}

\begin{aligned} \int \tan x d x = \int \frac{\sin x}{\cos x} d x = \int \frac{d (- \cos x)}{\cos x} = - \ln | \cos x | + C \\ \int \cot x d x = \int \frac{\cos x}{\sin x} d x = \int \frac{d (\sin x)}{\sin x} = \ln | \sin x | + C \end{aligned}

\begin{aligned} \int \sin^{2 k + 1} x \cos^{n} x d x & = \int (1 - \cos^{2} x)^{k} \cos^{n} x d (- \cos x) = - \int (1 - u^{2})^{k} u^{n} d u \\ \int \cos^{2 k + 1} x \sin^{n} x d x & = \int (1 - \sin^{2} x)^{k} \sin^{n} x d (\sin x) = \int (1 - u^{2})^{k} u^{n} d u \end{aligned}

\begin{array}{r} \int \sin^{2 k} x \cos^{2 l} x d x = \int {[\frac{1}{2} (1 - \cos 2 x)]}^{k} {[\frac{1}{2} (1 + \cos 2 x)]}^{l} d x \end{array}

\begin{array}{r} \int \tan^{n} x \sec^{2 k} x d x = \int \tan^{n} x (1 + \tan^{2} x)^{k - 1} d (\tan x) = \int u^{n} (1 + u^{2})^{(k - 1)} d u \end{array}

\begin{aligned} \int \tan^{2 k + 1} x \sec^{n + 1} x d x & = \int \tan^{2 k} x \sec^{n} x \tan x \sec x d x \\ = \int (\sec^{2} x - 1)^{k} \sec^{n} x d \sec x = \int (u^{2} - 1)^{k} u^{n} d u \end{aligned}

\begin{aligned} \int \sec x d x = \ln | \sec x + \tan x | + C \\ \int \csc x d x = \ln | \csc x - \cot x | + C \end{aligned}

也称为三角换元法，利用三角关系代换来换元：

\begin{array}{r} \int f (x) d x = {[\int f [ψ (t)] ψ^{'} (t) d t]}_{t = ψ^{- 1} (x)} \end{array}

令 $x = a \sin t \cos^{2} t + \sin^{2} t = 1$ 有：

\begin{aligned} \int \sqrt{a^{2} - x^{2}} d x & = \int a \cos t \cdot a \cos t d t = a^{2} \int \frac{1}{2} (1 + \cos 2 t) d t \\ = \frac{a^{2}}{2} t + \frac{a^{2}}{2} \sin t \cos t + C = \frac{a^{2}}{2} \arcsin \frac{x}{a} + \frac{1}{2} x \sqrt{a^{2} - x^{2}} + C \end{aligned}

令 $x = \tan t \tan^{2} t + 1 = \sec^{2} t$ 有：

\begin{aligned} \int \frac{d x}{\sqrt{x^{2} + a^{2}}} & = \int \frac{a \sec^{2} t}{a \sec t} d t = \int \sec t d t = \ln | \sec t + \tan t | + C \\ = \ln (\frac{x}{a} + \frac{\sqrt{x^{2} + a^{2}}}{a}) + C = \ln (x + \sqrt{x^{2} + a^{2}}) + C_{1} \end{aligned}

令 $x = \sec t \sec^{2} t - 1 = \tan^{2} t$ 有：

\begin{aligned} \int \frac{d x}{\sqrt{x^{2} - a^{2}}} & = \int \frac{a \sec t \tan t}{a \tan t} d t = \int \sec t d t = \ln (\sec t + \tan t) + C \\ = \ln (\frac{x}{a} + \frac{\sqrt{x^{2} - a^{2}}}{a}) + C = \ln (x + \sqrt{x^{2} - a^{2}}) + C_{1} \end{aligned}