微积分学/不定积分/三角代换法

$\begin{array}{c}\sqrt{a^2-u^2}u=a\sin \theta ,\frac{-\pi }{2}<\theta <\frac{\pi }{2}\\ \sqrt{a^2+u^2}u=a\tan \theta ,\frac{-\pi }{2}<\theta <\frac{\pi }{2}\\ \sqrt{u^2-a^2}u=a\sec \theta ,0<\theta <\pi ,\theta \ne \frac{\pi }{2}\end{array}$

例题

$\begin{array}{cc}\int \frac{1}{\sqrt{x^2+a^2}}dx& a>0\\ x=a\tan \theta & dx=a{\sec }^2\theta d\theta \\ \int \frac{1}{\sqrt{x^2+a^2}}dx& =\int \frac{a{\sec }^2\theta }{\sqrt{a^2{\tan }^2\theta +a^2}}d\theta =\int \sec \theta d\theta =\ln (\tan \theta +\sec \theta )+\text{C}\\ & \tan \theta =\frac{x}{a}\sec \theta =\sec \left({\tan }^{-1}\frac{x}{a}\right)=\pm \sqrt{{\tan }^2\theta +1}=\pm \sqrt{{\left(\frac{x}{a}\right)}^2+1}\\ & =\ln (\frac{x}{a}+\sqrt{\frac{x^2}{a^2}+1})+\text{C}=\ln (\frac{x}{a}+\frac{1}{a}\sqrt{x^2+a^2})+\text{C}\\ & =\ln \left(\frac{x+\sqrt{x^2+a^2}}{a}\right)+\text{C}\end{array}$