查看“︁微积分学/不定积分/三角代換法”︁的源代码

<math>\begin{align}\sqrt{a^2-u^2}\qquad u=a\sin\theta,\,\frac{-\pi}2<\theta<\frac\pi2\\\sqrt{a^2+u^2}\qquad u=a\tan\theta,\,\frac{-\pi}2<\theta<\frac\pi2\\\sqrt{u^2-a^2}\qquad u=a\sec\theta,\,0<\theta<\pi,\,\theta\ne\frac\pi2\end{align}</math>
==例題==
<math>\begin{align}\int\frac1{\sqrt{x^2+a^2}}dx&\qquad a>0\\x=a\tan\theta&\qquad dx=a\sec^2\theta\,d\theta\\\int\frac1{\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\qquad\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\left(\frac x a+\sqrt{\frac{x^2}{a^2}+1}\right)+\text{C}=\ln\left(\frac x a+\frac1 a\sqrt{x^2+a^2}\right)+\text{C}\\&=\ln\left(\frac{x+\sqrt{x^2+a^2}}a\right)+\text{C}\end{align}</math>