第二类换元法之其它方法

笔记下载:第二类换元法的其它方法

1,双曲代换

\(\qquad x=sin ht\),\(dx=\cos htdt\)

或\(\quad x=\cos ht\),\(dx=\sin htdt\)

双曲函数:\(\displaystyle \sin hx=\frac{e^x-e^{-x}}{2}\),\(\displaystyle \cos hx=\frac{e^x+e^{-x}}{2}\)

\(\displaystyle\qquad\ \ \Longrightarrow \cos h^2x-1=\sin h^2x\)

反双曲函数:\(\displaystyle \sin h^{-1}x=\ln \left(x+\sqrt{x^2+1}\right)\),\(\displaystyle \cos h^{-1}x=\ln \left(x+\sqrt{x^2-1}\right)\)

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

解答:令\(x=a\cos ht\),\(dx=a\sin htdt\)

则\(\displaystyle\quad \int \frac{dx}{\sqrt{x^2-a^2}}=\int \frac{a\sin htdt}{\sqrt{a^2\cosh^2t-a^2}}\)

\(\displaystyle\qquad\qquad\qquad\quad=\int \frac{\sin ht}{\sin ht}dt=\int dt=t+C\)

回代:\(\displaystyle x=a\cos ht\quad\Rightarrow\quad\cos ht=\frac{x}{a}\)

\(\displaystyle\qquad\qquad\qquad\qquad\quad\Rightarrow\quad t=\cos h^{-1} \frac{x}{a}=\ln|\frac{x}{a}+\sqrt{\left(\frac{x}{a}\right)^2-1}|\)

因此,\(\displaystyle\int \frac{dx}{\sqrt{x^2-a^2}}=\ln|\frac{x}{a}+\sqrt{\left(\frac{x}{a}\right)^2-1}|+C\)

\(\displaystyle\qquad\qquad\qquad\qquad \ =\ln|\frac{x}{a}+\frac{1}{a}\sqrt{x^2-a^2}|+C\)

\(\displaystyle\qquad\qquad\qquad\qquad\ =\ln|x+\sqrt{x^2-a^2}|+C\)

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

解答:令\(\quad x=a\sin ht\),\(dx=a\cos htdt\),

则\(\displaystyle\qquad\quad\int\frac{dx}{\sqrt {x^2+a^2}}\)

\(\displaystyle\qquad\ \ =\int \frac{a\cos htdt}{\sqrt {a^2\sin h^2t+a^2}}\)

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

\(\displaystyle\qquad\ \ =\sin h^{-1} \frac{x}{a}+C\)

\(\displaystyle\qquad\ \ =\ln|x+\sqrt{x^2+a^2}|+C\)

2,例代换:\(\displaystyle\quad x=\frac{1}{t}\),\(\displaystyle\quad dx=-\frac{1}{t^2}dt\)

例3,求\(\displaystyle\quad\int \frac{\sqrt{a^2-x^2}}{x^4}dx\)

解答:令\(\displaystyle\quad x=\frac{1}{t}\),\(\displaystyle\quad dx=-\frac{1}{t^2}dt\)

则\(\displaystyle\quad\int \frac{\sqrt{a^2-x^2}}{x^4}dx\)

\(\displaystyle\quad=\int \frac{\sqrt{a^2-\frac{1}{t^2}}}{\frac{1}{t^4}}\cdot \left(-\frac{1}{t^2}\right) dt\)

\(\displaystyle\quad=-\int t^2\cdot \sqrt{a^2-\frac{1}{t^2}}dt\)

由于\(\displaystyle\quad t>0,\quad x>0\)

接\(\displaystyle\quad=-\int t^2\cdot \sqrt{a^2-\frac{1}{t^2}}dt\)

\(\displaystyle\qquad=-\int t\sqrt{a^2t^2-1} dt\)

令\(\displaystyle\quad u=a^2t^2-1\),则\(\displaystyle\quad du=2a^2tdt\)

接\(\displaystyle\qquad=-\int t\sqrt{a^2t^2-1} dt\)

\(\displaystyle\qquad=-\frac{1}{2a^2}\int \sqrt{u}du\)

\(\displaystyle\qquad=-\frac{1}{2a^2}\cdot \frac{2}{3}u^{\frac{3}{2}}+C\)

\(\displaystyle\qquad=-\frac{1}{3a^2}\left(a^2t^2-1\right)^{\frac{3}{2}}+C\)

\(\displaystyle\qquad=-\frac{1}{3a^2}\left(\frac{a^2}{x^2}-1\right)^{\frac{3}{2}}+C\)

\(\displaystyle\qquad=-\frac{1}{3a^2x^3}\left(a^2-x^2\right)^{\frac{3}{2}}+C\)

3,万能代换:\(\displaystyle\quad x=\tan \frac{t}{2}\),\(\displaystyle\quad t=\tan \frac{x}{2}\)

\(\displaystyle\quad\sin x=\frac{2\tan \frac{x}{2}}{1+\tan \frac{x}{2}}=\frac{2t}{1+t^2}\)

\(\displaystyle\quad\cos x=\frac{1-\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}=\frac{1-t^2}{1+t^2}\)

\(\displaystyle\quad f(\sin x, \cos x)=f(\frac {2t}{1+t^2}, \frac{1-t^2}{1+t^2})\quad\Longrightarrow\quad\) 有理函数