第一类换元法

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\(du=(\cos x+\sin x)dx\)

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