Integration By Substitution
说明
一、第一类换元法
核心是通过凑微分来换元：
\int f\left[\varphi(x)\right]\varphi'(x)\; \mathrm{d}x=\left[\int f(u)\; \mathrm{d}u\right]_{u=\varphi(x)} \end{align}$$ $$\begin{align} \int \dfrac{1}{a^{2}+x^{2}}\; \mathrm{d}x=\dfrac{1}{a} \int \dfrac{1}{1+(\dfrac{x}{a})^{2}}\; \mathrm{d} \dfrac{x}{a}=\dfrac{1}{a} \arctan \dfrac{x}{a}+C \end{align}$$ $$\begin{align} \int \dfrac{\mathrm{d}x}{\sqrt{ a^{2}-x^{2} }} =\int \dfrac{ \mathrm{d} \dfrac{x}{a}}{\sqrt{ 1-(\dfrac{x}{a})^{2} }}=\arcsin \dfrac{x}{a}+C \end{align}$$ $$\begin{align} \int \dfrac{1}{x^{2}-a^{2}}\, dx =\int \dfrac{1}{2a}(\dfrac{1}{x-a}-\dfrac{1}{x+a})\, dx =\dfrac{1}{2a}\ln | \dfrac{x-a }{x+a}|+C \end{align}$$ $$\begin{align} & \int \tan x\, dx =\int \dfrac{\sin x}{\cos x}\; \mathrm{d}x= \int \dfrac{\mathrm{d}(-\cos x)}{\cos x} =-\ln |\cos x|+C \\ \\ & \int \cot x\, dx= \int \dfrac{\cos x}{\sin x}\; \mathrm{d}x=\int \dfrac{\; \mathrm{d}(\sin x)}{\sin x} =\ln |\sin x|+C \end{align}$$ $$\begin{align} \int \sin ^{2k+1}x\;\cos ^{n}x \; \mathrm{d}x & =\int (1-\cos ^{2} x)^{k}\cos ^{n}x\; \mathrm{d}(-\cos x)=-\int (1-u^{2})^{k}u^{n}\; \mathrm{d}u \\ \\ \int \cos ^{2k+1}x\;\sin ^{n}x \; \mathrm{d}x & =\int (1-\sin ^{2} x)^{k}\sin ^{n}x\; \mathrm{d}(\sin x)=\int (1-u^{2})^{k}u^{n}\; \mathrm{d}u \end{align}$$ $$\begin{align} \int \sin ^{2k}x \cos ^{2l}x \; \mathrm{d}x=\int \left[\dfrac{1}{2}(1-\cos{2}x)\right]^{k} \left[\dfrac{1}{2}(1+\cos{2}x)\right]^{l} \; \mathrm{d}x \end{align}$$ $$\begin{align} \int \tan ^{n}x\sec ^{2k}x\; \mathrm{d}x=\int \tan ^{n}x(1+\tan ^{2}x)^{k-1}\; \mathrm{d}(\tan x)=\int u^{n}(1+u^{2})^{(k-1)}\; \mathrm{d}u \end{align}$$ $$\begin{align} \int \tan ^{2k+1}x\sec ^{n+1}x\; \mathrm{d}x & =\int \tan ^{2k}x\sec ^{n}x \tan x\sec x\; \mathrm{d}x \\ & =\int (\sec ^{2}x-1)^{k}\sec ^{n}x\; \mathrm{d}\sec x=\int(u^{2}-1)^{k}u^{n}\; \mathrm{d}u \end{align}$$ $$\begin{align} & \int \sec x\, dx =\ln |\sec x+\tan x|+C \\ \\ & \int \csc x\, dx =\ln |\csc x-\cot x|+C \end{align}$$ ### 二、第二类换元法 也称为三角换元法，利用[[三角公式\|三角关系]]代换来换元： $$\begin{align} \int f(x)\; \mathrm{d}x =\left[\int f[\psi(t)] \psi'(t)\; \mathrm{d}t \right]_{t=\psi^{-1}(x)} \end{align}$$  令 $x=a\sin t\quad \cos ^{2}t+\sin ^{2}t=1$ 有： $$\begin{align} \int \sqrt{ a^{2}-x^{2} }\; \mathrm{d}x & =\int a\cos t\cdot a\cos t \; \mathrm{d}t=a^{2}\int \dfrac{1}{2}(1+\cos2t)\; \mathrm{d}t \\ & =\dfrac{a^{2}}{2}t+ \dfrac{a^{2}}{2}\sin t\cos t+C=\dfrac{a^{2}}{2}\arcsin \dfrac{x}{a}+\dfrac{1}{2}x\sqrt{ a^{2}-x^{2} }+C \end{align}$$ 令 $x=\tan t\quad \tan ^{2}t+1=\sec ^{2}t$ 有： $$\begin{align} \int \dfrac{\mathrm{d}x}{\sqrt{ x^{2}+a^{2} }} & =\int \dfrac{a\sec ^{2} t}{a\sec t}\; \mathrm{d}t=\int\sec t\; \mathrm{d}t=\ln \left\lvert \sec t+\tan t \right\rvert+C \\ & =\ln \left(\dfrac{x}{a}+\dfrac{\sqrt{ x^{2}+a^{2} }}{a}\right)+C=\ln (x+\sqrt{ x^{2}+a^{2} })+C_{1} \end{align}$$ 令 $x=\sec t\quad \sec ^{2}t-1=\tan ^{2}t$ 有： $$\begin{align} \int \dfrac{\mathrm{d}x}{\sqrt{ x^{2}-a^{2} }} & =\int \dfrac{a\sec t \tan t }{a\tan t}\; \mathrm{d}t=\int \sec t\; \mathrm{d}t =\ln (\sec t+\tan t)+C \\ & =\ln \left(\dfrac{x}{a}+\dfrac{\sqrt{ x^{2}-a^{2} }}{a}\right)+C=\ln (x+\sqrt{ x^{2}-a^{2} }) +C_{1} \end{align}$$