前置知识:第二类换元法
题1: 计算∫1(a2−x2)32dx\int \dfrac{1}{(a^2-x^2)^{\frac 32}}dx∫(a2−x2)231dx
解:
\qquad令x=asintx=a\sin tx=asint,t=arcsinxat=\arcsin \dfrac xat=arcsinax,dx=acostdtdx=a\cos tdtdx=acostdt
\qquad原式=∫1(acost)3⋅acostdt=1a2∫1cos2tdt=\int\dfrac{1}{(a\cos t)^3}\cdot a\cos tdt=\dfrac{1}{a^2}\int\dfrac{1}{\cos^2 t}dt=∫(acost)31⋅acostdt=a21∫cos2t1dt
=1a2tant+C=xa2a2−x2+C\qquad\qquad =\dfrac{1}{a^2}\tan t+C=\dfrac{x}{a^2\sqrt{a^2-x^2}}+C=a21tant+C=a2a2−x2x+C
题2: 计算∫1x21+x2dx\int\dfrac{1}{x^2\sqrt{1+x^2}}dx∫x21+x21dx
解:
\qquad令x=tantx=\tan tx=tant,t=arctanxt=\arctan xt=arctanx,dx=1cos2tdtdx=\dfrac{1}{\cos^2 t}dtdx=cos2t1dt
\qquad原式=∫1tan2t⋅1cost⋅1cos2tdt=∫1tan2tcostdt=\int\dfrac{1}{\tan^2 t\cdot \frac{1}{\cos t}}\cdot \dfrac{1}{\cos^2 t}dt=\int\dfrac{1}{\tan^2 t\cos t}dt=∫tan2t⋅cost11⋅cos2t1dt=∫tan2tcost1dt
=∫1tant⋅1sintdt=∫cottcsctdt\qquad\qquad =\int \dfrac{1}{\tan t}\cdot\dfrac{1}{\sin t}dt=\int \cot t \csc tdt=∫tant1⋅sint1dt=∫cottcsctdt
=−csct+C=−1+cot2t+C\qquad\qquad =-\csc t+C=-\sqrt{1+\cot^2 t}+C=−csct+C=−1+cot2t+C
=−1+1x2+C=−x2+1x+C\qquad\qquad =-\sqrt{1+\frac{1}{x^2}}+C=-\dfrac{\sqrt{x^2+1}}{x}+C=−1+x21+C=−xx2+1+C
题3: 计算∫1x2−a2dx\int \dfrac{1}{\sqrt{x^2-a^2}}dx∫x2−a21dx
解:
\qquad令x=asectx=a\sec tx=asect,dx=asecttantdtdx=a\sec t\tan tdtdx=asecttantdt
\qquad原式=∫1tant⋅secttantdt=∫sectdt=\int\dfrac{1}{\tan t}\cdot \sec t\tan tdt=\int\sec tdt=∫tant1⋅secttantdt=∫sectdt
=ln∣sect+tant∣+C=ln∣x+x2−a2a∣+C\qquad\qquad =\ln|\sec t+\tan t|+C=\ln|\dfrac{x+\sqrt{x^2-a^2}}{a}|+C=ln∣sect+tant∣+C=ln∣ax+x2−a2∣+C
要学会观察被积函数,遇到有根号且根号的形式为类似a2±x2\sqrt{a^2\pm x^2}a2±x2或x2±a2\sqrt{x^2\pm a^2}x2±a2的,一般都可以用三角代换来做。