第二类换元法之三角代换

笔记下载:第二类换元法之三角代换

1,第一类换元:\(u=g(x)\qquad\int f(g|x|)g'(x)dx=\int f(u)du\)

2,第二类换元:\(x=\phi(t)\qquad\int f(x)dx=\int f(\phi(t))\phi(t)dt\)

3,三角代换:

(1)\(\sqrt{a^2-x^2}\),令\(x=a\sin t\),\(\sqrt{a^2-x^2}=\sqrt{a^2-a^2\sin^2 t}=a\cos t\),\(dx=a\cos tdt\)

(2)\(\sqrt{a^2+x^2}\),令\(x=a\tan t\),\(\left(1+\tan^2 t=\sec^2 t\right)\Longrightarrow\sqrt{a^2+x^2}=a\sec t\)

(3)\(\sqrt{x^2-a^2}\),令\(x=a\sec t\),\(\left(1+\tan^2 t=\sec^2 t\right)\Longrightarrow\sqrt{x^2-a^2}=a\tan t\)

例1,求\(\displaystyle\int \frac{\sqrt{9-x^2}}{x^2}dx\)

解答:令\(x=3\sin t\),则\(dx=3\cos tdt\)

\(\displaystyle\int \frac{\sqrt{9-x^2}}{x^2}dx=\int \frac{\sqrt{9-9\sin^2 t}}{9\sin^2 t}\cdot\cos tdt=\int \frac{3\cos t\cdot 3\cos t}{9 sin^2 t}dt=\int \frac{\cos^2 t}{\sin^2 t}dt=\int \cot^2 tdt\)

又因\(\cot^2 t+1=\csc^2t\)

\(\int \cot^2 tdt=\int (csc^2 t-1)dt=-\cot t-t+C\)

要换回\(x\),利用直角三角形,\(x=3\sin t\)

\(\displaystyle\Rightarrow\sin t=\frac{x}{3}\quad\Rightarrow\quad t=\arcsin\frac{x}{3}\quad\Rightarrow\quad\cot t=\frac{\sqrt{9-x^2}}{x}\)

所以,\(\displaystyle\int \frac{\sqrt{9-x^2}}{x^2}dx=-\frac{9-x^2}{x}-\arcsin\frac{x}{3}+C\)

例2,求\(\displaystyle\int \frac{dx}{x^2\sqrt{x^2+4}}\)

解答:令\(x=2\tan t\),则\(dx=2\sec^2 tdt\)

\(\displaystyle\int \frac{dx}{x^2\sqrt{x^2+4}}=\int \frac{2\sec^2 tdt}{4\tan^2t\sqrt{4tan^2\cdot t\cdot\sec t}}=\int \frac{\sec t}{4\tan^2 t}dt\)

\(\displaystyle=\frac{1}{4}\int \frac{1}{\cos t}\cdot\frac{cos^2 t}{\sin^2 t}dt=\frac{1}{4}\int \frac{\cos t}{\sin^2 t}dt=\frac{2}{4}\int \cot t\cdot\csc tdt=\frac{1}{4}\csc t+C\)

回代:\(\displaystyle x=2\tan t\quad\Rightarrow\quad\tan t=\frac{x}{2}\quad\Rightarrow\quad\csc t=\frac{\sqrt{4+x^2}}{x}\)

\(\displaystyle\Rightarrow\quad\int \frac{dx}{x^2\sqrt{x^2+4}}=-\frac{1}{4}\cdot\frac{\sqrt{4+x^2}}{x}+C\)

例3,求\(\displaystyle\int \frac{dx}{\sqrt{x^2-a^2}}\)

解答:令\(x=a\sec t\),则\(dx=a \sec t\tan t dt\)

\(\displaystyle\int \frac{dx}{\sqrt{x^2-a^2}}=\int \frac{a \sec t\tan t dt}{\sqrt {a^2 \sec^2 t-a^2}}=\frac{1}{a}\int \frac{a \sec t\tan t}{\tan t}dt\)

\(\displaystyle\qquad\ \ =\int \sec tdt=\int \frac{1}{\sin t}dt\)

\(\displaystyle\qquad\ \ =\int \frac{\sin t}{sin^2 t}dt=\int \frac{\sin tdt}{1-\cos^2 t}\)

令\(u=\cos t\),则\(du=\sin tdt\)

接\(\displaystyle\int \frac{\sin tdt}{1-\cos^2 t}=-\int \frac{du}{1-u^2}=-\frac{1}{2}\int \left(\frac{1}{1-u}-\frac{1}{i+u}\right)du\)

 \(\displaystyle\qquad\ \ =-\frac{1}{2}\left(-\ln|1-n|-\ln|1+n|\right)+C=\frac{1}{2}\ln |1-n^2|+C\)

 \(\displaystyle\qquad\ \ =\frac{1}{2}\ln|\sin^2 t|+C=\ln |\sin t|+C\)

又因\(\displaystyle\frac{1}{1-u^2}=1-\frac{1}{(1-u)(1+u)}=\frac{A}{1-u}+\frac{B}{1+u}\),

得\(A(1+u)+B(1-u)=1\),

则\(\quad\Rightarrow\quad u=1\qquad\Rightarrow\quad 2A=1\quad A=\frac{1}{2}\)

\(\qquad\qquad u=-1\qquad\Rightarrow\quad -2B=1\quad B=-\frac{1}{2}\)

回代:令\(x=a\sec t\),则\(\displaystyle\sec t=\frac{x}{a}\),\(\displaystyle\sin t=\frac{\sqrt{x^2-a^2}}{x}\)

\(\displaystyle\Longrightarrow\int \frac{dx}{\sqrt{x^2-a^2}}=\ln|\frac{x^2-a^2}{x}|+C\)

例4,求\(\displaystyle\int \frac{dx}{\sqrt{1+x-a^2}}\)

解答:先把\(\sqrt{ax^2+bx+c}\)配方,

\(\displaystyle\int \frac{dx}{\sqrt{1+x-a^2}}=\int \frac{dx}{\sqrt{-(x^2-x)}}+1=\int \frac{dx}{\sqrt{-(x-\frac{1}{2})^2+\frac{1}{4}+1}}\)

\(\displaystyle\qquad=\int \frac{dx}{\sqrt {\frac{5}{4}-(x-\frac{1}{2})^2}}=\int \frac{dx}{\sqrt {\left(\frac{\sqrt 5}{2}\right)^2-(x-\frac{1}{2})^2}}\)

令\(\displaystyle x-\frac{1}{2}=\frac {\sqrt 5}{2}\sin t\)

所以,原式\(=\displaystyle\int \frac{\frac{\sqrt 5}{2}\cos tdt}{\sqrt {\left(\frac{\sqrt 5}{2}\right)^2-\left(\frac{\sqrt 5}{2}\right)^2\sin^2 t}}=\int \frac{\frac{\sqrt 5}{2}\cos tdt}{\frac{\sqrt 5}{2}\cos t}\)

\(\displaystyle\qquad\ \ =\int dt=t+C\)

回代:\(\displaystyle\sin t=\frac{2(x-\frac{1}{2})}{\sqrt 5}\),则\(\displaystyle t=\arcsin \left(\frac{2x-1}{\sqrt 5}\right)\)

最后,\(\displaystyle\int \frac{dx}{\sqrt{1+x-a^2}}=\arcsin \frac{2x-1}{\sqrt 5}+C\)