第一换元积分

第一换元积分（凑积分）

\[设f(\mu)具有原函数F(\mu),\mu=\varphi(x)可导，则有换元公式\\ \begin{align} \int f[\varphi(x)]\varphi'(x)dx&=\int f[\varphi(x)]d\varphi(x)=^{u=\varphi(x)}\int f)(\mu)d\mu\\ &=F(\mu)+C=F[\varphi(x)]+C \end{align} \]

常见的凑微分的形式

\[\begin{align} &\int f(ax^n+b)x^{n-1}=\frac1{na}\int f(ax^n+b)(a\neq0,n\neq0)\\ &\int f(\sin x)\cos x dx=\int f(\sin x)d(\sin x)\\ &\int f(\cos x)\sin xdx=-\int f(\cos x)d(\cos x)\\ &\int f(\ln x)\frac{dx}x=\int f(\ln x)d(\ln x)\\ &\int f(\sqrt x)\frac{dx}{\sqrt x}=2\int f(\sqrt x)d(\sqrt x)\\ &\int f(\frac1x)\frac{dx}{x^2}=-\int f(\frac1x)d(\frac1x)\\ &\int f(\arctan x)\frac{dx}{1+x^2}=\int f(\arctan x)d(\arctan x)\\ &\int f(\arcsin x)\frac{dx}{\sqrt{1+x^2}}=\int f(\arcsin x)d\arcsin x \end{align} \]

\[【例】求\int 2\cos 2xdx\\ \begin{align} =&\int f(\cos 2x)d2x\\ =&\sin x+C \end{align} \]

\[【例】求\int \frac1{a^2+x^2}dx(a\neq0)\\ \begin{align}&=\int \frac1{a^2(1+\frac {x^2}{a^2})}dx\\ &=\frac1a\int \frac1{1+(\frac x{a})^2}d\frac xa\\ &=\frac1a\arctan \frac xa+C \end{align} \]

\[【例】\int \frac{\sin x \cos x}{1+\sin^4x}dx\\ \begin{align} &=\int \frac{\sin x}{1+\sin ^4x}d\sin x\\ &=\frac12\int\frac1{1+(\sin^2)^2}d\sin ^2x\\ &=\frac12\arctan (\sin ^2x)+C \end{align} \]

\[【例】求\int \sin^3xdx\\ =\int \sin^2\cdot \sin xdx\\ =\int(1-\cos^2x)\cdot \sin xdx\\ =\int (1-\cos^2x)d(-\cos x)\\ =\frac{\cos ^3x}3-\cos x+C \]

\[【例】求\int \tan xdx\\ =\int \frac{\sin x}{\cos x}dx=-\int \frac1{\cos x}d(\cos x)\\ =-\ln|\cos x|+C \]

\[【例】求\int \cos ^2xdx\\ =\int \frac{1+\cos 2x}{2}dx\\ =\frac12\int 1dx+\frac14\int \cos 2x d2x\\ =\frac 12x + \frac14\sin 2x+C \]

\[【例】求\int \sec ^6xdx\\ =\int \sec^4\cdot \sec^2dx\\ =\int (1+\tan^2)^2.\sec^2dx\\ =\int (1+\tan^2)^2d\tan x\\ =C+\tan x+\frac{2\tan^3 x}3+\frac{\tan^5x}5 \]

\[【例】求\int \csc xdx\\ =\int \frac1{\sin x}dx=\int \frac1{\sin \frac x2\cos \frac x2}d\frac x2\\ =\int \frac1{\tan \frac x 2\cdot \cos^2 \frac x2}d\frac x2=\int \frac1{\tan \frac x2}\cdot\sec ^2\frac x2dx\\ =\int \frac1{\tan \frac x2}d\tan \frac x2\\ =\ln|\tan \frac x2|+C\\ \tan \frac x2=\frac{\sin \frac x2}{\cos \frac x2}=\frac {2\sin \frac x2\cdot \sin \frac x2}{2\sin \frac x2\cdot \cos \frac x2}\\ =\frac{2\sin ^2\frac x2}{\sin x}=\frac{1-\cos x}{\sin x}=\csc x- \cot x \]

\[\ln|\tan \frac x2|+C=\ln|\csc x -\cot x|+C \]

\[【例】求\int \sec xdx\\ =\int \frac 1{\cos x}dx=\int \frac{1}{\sin (\frac \pi 2+x)}d_{\frac \pi 2+x}=^{u=x+\frac \pi2}\\ \int \frac 1{\sin u}du=\int \csc udu\\ =\ln|\csc u-\cot u|+C \]