笔记下载：第一类换元法
1，令 \(u=g(x)\)，则\(du=g'(x)dx\)
\(\displaystyle\int f\left(g(x)\right) \cdot g'(x) dx=\int f(u)du\)
例1，求 \(\displaystyle\int e^{2x}dx\)
解答：令 \(u=2x\)，则\(du=2dx\)，\(dx=\frac{1}{2}du\)
\(\displaystyle\int e^{2x}dx=\int e^u\cdot \frac{1}{2} du=\frac{1}{2}e^u+C=\frac{1}{2}e^{2x}+C\)
例2，求\(\displaystyle\int \frac{dx}{x+4}\)
解答：令\(u=x+4\)，则\(du=dx\)
\(\displaystyle\int \frac{dx}{x+4}=\int \frac{1}{u}du=\ln|u|+C=\ln|x+4|+C\)
例3，求\(\displaystyle\int \frac{dx}{ax+b}\)
解答：令\(u=ax+b\)，则\(du=adx\)，\(\displaystyle dx=\frac{1}{a} dn\)
\(\displaystyle\int \frac{dx}{ax+b}=\int \frac{1}{u}\cdot\frac{1}{a}du=\frac{1}{a}\int \frac{1}{u}du=\frac{1}{a}\ln|a|+C\)
例4，求\(\displaystyle\int 2xe^{x^2}dx\)
解答：令\(u=x^2\)，则\(du=2xdx\)
\(\displaystyle\int 2xe^{x^2}dx=\int e^udu=e^u+C=e^{x^2}+C\)
例5，求\(\displaystyle\int e^x\cdot \cos (e^x) dx\)
解答：令\(u=e^x\)，则\(du=e^xdx\)
\(\displaystyle\int e^x\cdot \cos (e^x)dx=\int \cos u du=\sin u+C\)
例6，求\(\displaystyle\int \frac{dx}{9+x^2}\)
解答：\(\displaystyle\quad=\frac{1}{9}\int \frac{dx}{1+\frac{x^2}{9}}=\frac{1}{9}\int \frac{dx}{1+(\frac{x}{3})^2}\)
令\(\displaystyle u=\frac{x}{3}\)，则\(\displaystyle du=\frac{1}{3}dx\)，\(\displaystyle dx=3du\)
接 \(\displaystyle\frac{1}{9}\int \frac{dx}{1+(\frac{x}{3})^2}=\frac{1}{3}\int \frac{du}{1+u^2}=\frac{1}{3}\arctan u+C=\frac{1}{3}\arctan \frac{x}{3}+C\)
例7，求\(\displaystyle\int \frac{dx}{\sqrt{a^2-x^2}}\)
解答：\(\displaystyle\int \frac{dx}{\sqrt{a^2-x^2}}=\frac{1}{a}\int \frac{dx}{\sqrt{1-\left(\frac{x}{a}\right)^2}}\)
令\(\displaystyle u=\frac{x}{a}\)，则\(dx=adu\)
接\(\displaystyle \frac{1}{a}\int \frac{dx}{\sqrt{1-\left(\frac{x}{a}\right)^2}}=\frac{1}{a}\int \frac{adu}{\sqrt{1-u^2}}=\arctan a+C=\arctan \frac{s}{a}+C\)
例8，求\(\displaystyle\int \tan^3 x\cdot \left(\sec x\right)^2 dx\)
解答：令\(u=\tan x\)，则\(du=\sec^2 xdx\)
\(\displaystyle\int \tan^3 x\cdot \sec^2 x dx=\int u^3du=\frac{1}{4}u^4+C=\frac{1}{4}\tan^4 x+C\)
例9，求\(\displaystyle\int \frac{x}{1+x^2}dx\)
解答：令\(u=x^2+1\)，则\(du=2xdx\)，\(xdx=\frac{1}{2}du\)
\(\displaystyle\int \frac{x}{1+x^2}dx=\frac{1}{2}\int \frac{1}{u}du=\frac{1}{2}\ln u+C=\frac{1}{2}ln(1+x^2)+C\)
例10，求\(\displaystyle\int \frac{dx}{x\ln x\cdot\ln(\ln x)}\)
解答：令\(u=\ln(\ln x)\)，则\(\displaystyle du=\frac{dx}{x\ln x}\)，
\(\displaystyle\int \frac{dx}{x\ln x\cdot\ln(\ln x)}=\int \frac{du}{u}=\ln |u|+C=\ln |\ln(\ln x)|+C\)
例11，求\(\displaystyle\int \sin x\cos x dx\)
解答：令\(u=\sin x)\)，则\(du=\cos xdx\)
\(\displaystyle\int \sin x\cos x dx=\int udu=\frac{1}{2}u^2+C=\frac{1}{2}\sin^2 x+C=\frac{1}{2}\int \sin (2x)dx=-\frac{1}{4}\cos (2x)+C\)
例12，求\(\int \sin^3 xdx\)
解答：\(\int \sin^3 xdx=\int \sin^2 x\sin x dx=\int (1-\cos^2 x)\sin xdx\)
令\(u=\cos x\)，则\(du=-\sin xdx\)
\(\displaystyle\int (1-\cos^2 x)\sin xdx=-\int (1-u^2)du=\frac{1}{3}u^3-u+C=\frac{1}{3}\cos^3 x-\cos x+C\)
2，第一类换元法的原则：哪部分最麻烦就换哪一部分
例13，求\(\displaystyle\int \frac{\sin x+\cos x}{\sqrt[3]{\sin x-\cos x}}dx\)
解答：令\(u=\sin x-\cos x\)，则\(du=(\cos x+\sin x)dx\)
\(\displaystyle\int \frac{\sin x+\cos x}{\sqrt[3]{\sin x-\cos x}}dx=\int \frac{du}{\sqrt[3]u}=\int u^{-\frac{1}{3}}du=\frac{3}{2}u^{\frac{2}{3}}+C=\frac{2}{3}(\sin x-\cos x)^{\frac{2}{3}}+C\)