# 第一类换元法

1，第一类换元法的公式：令 $$u=g(x)$$，则$$du=g'(x)dx$$

$\int f\left(g(x)\right) \cdot g'(x) dx=\int f(u)du$

$\int e^{2x}dx=\int e^u\cdot \frac{1}{2} du=\frac{1}{2}e^u+C=\frac{1}{2}e^{2x}+C$

$\int \frac{dx}{x+4}=\int \frac{1}{u}du=\ln|u|+C=\ln|x+4|+C$

$\int \frac{dx}{ax+b}=\int \frac{1}{u}\cdot\frac{1}{a}du=\frac{1}{a}\int \frac{1}{u}du=\frac{1}{a}\ln|a|+C$

$\int 2xe^{x^2}dx=\int e^udu=e^u+C=e^{x^2}+C$

$\int e^x\cdot \cos (e^x)dx=\int \cos u du=\sin u+C$

$\int \frac{dx}{9+x^2}=\frac{1}{9}\int \frac{dx}{1+\frac{x^2}{9}}=\frac{1}{9}\int \frac{dx}{1+(\frac{x}{3})^2}$

\begin{align*}\int \frac{dx}{9+x^2}&=\frac{1}{9}\int \frac{dx}{1+(\frac{x}{3})^2}=\frac{1}{3}\int \frac{du}{1+u^2}\\ &=\frac{1}{3}\arctan u+C=\frac{1}{3}\arctan \frac{x}{3}+C\end{align*}

$\int \frac{dx}{\sqrt{a^2-x^2}}=\frac{1}{a}\int \frac{dx}{\sqrt{1-\left(\frac{x}{a}\right)^2}}$

\begin{align*} \int \frac{dx}{\sqrt{a^2-x^2}}&=\frac{1}{a}\int \frac{dx}{\sqrt{1-\left(\frac{x}{a}\right)^2}}=\frac{1}{a}\int \frac{adu}{\sqrt{1-u^2}}\\ &=\arcsin u+C=\arcsin \frac{x}{a}+C\end{align*}

$\int \tan^3 x\cdot \sec^2 x dx=\int u^3du=\frac{1}{4}u^4+C=\frac{1}{4}\tan^4 x+C$

$\int \frac{x}{1+x^2}dx=\frac{1}{2}\int \frac{1}{u}du=\frac{1}{2}\ln u+C=\frac{1}{2}ln(1+x^2)+C$

$\int \frac{dx}{x\ln x\cdot\ln(\ln x)}=\int \frac{du}{u}=\ln |u|+C=\ln |\ln(\ln x)|+C$

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

$\int \sin^3 xdx=\int \sin^2 x\sin x dx=\int (1-\cos^2 x)\sin xdx$

$\int (1-\cos^2 x)\sin xdx=-\int (1-u^2)du=\frac{1}{3}u^3-u+C=\frac{1}{3}\cos^3 x-\cos x+C$

2，我们说过第一类换元法的原则：哪部分最麻烦就换哪一部分。再来看一个比较特殊的例子，虽然不能明确知道换元以后是什么样，但是换了以后，就看出换元以后对积分的简化了。

\begin{align*}\int \frac{\sin x+\cos x}{\sqrt[3]{\sin x-\cos x}}dx&=\int \frac{du}{\sqrt[3]u}=\int u^{-\frac{1}{3}}du=\frac{3}{2}u^{\frac{2}{3}}+C\\ &=\frac{2}{3}(\sin x-\cos x)^{\frac{2}{3}}+C\end{align*}