# 微积分学/不定积分/三角代換法

{\displaystyle {\begin{aligned}{\sqrt {a^{2}-u^{2}}}\qquad u=a\sin \theta ,\,{\frac {-\pi }{2}}<\theta <{\frac {\pi }{2}}\\{\sqrt {a^{2}+u^{2}}}\qquad u=a\tan \theta ,\,{\frac {-\pi }{2}}<\theta <{\frac {\pi }{2}}\\{\sqrt {u^{2}-a^{2}}}\qquad u=a\sec \theta ,\,0<\theta <\pi ,\,\theta \neq {\frac {\pi }{2}}\end{aligned}}}

## 例題

{\displaystyle {\begin{aligned}\int {\frac {1}{\sqrt {x^{2}+a^{2}}}}dx&\qquad a>0\\x=a\tan \theta &\qquad 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}}\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}}+{\frac {1}{a}}{\sqrt {x^{2}+a^{2}}}\right)+{\text{C}}\\&=\ln \left({\frac {x+{\sqrt {x^{2}+a^{2}}}}{a}}\right)+{\text{C}}\end{aligned}}}