a 2 − u 2 u = a sin θ , − π 2 < θ < π 2 a 2 + u 2 u = a tan θ , − π 2 < θ < π 2 u 2 − a 2 u = a sec θ , 0 < θ < π , θ ≠ π 2 {\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}}}
∫ 1 x 2 + a 2 d x a > 0 x = a tan θ d x = a sec 2 θ d θ ∫ 1 x 2 + a 2 d x = ∫ a sec 2 θ a 2 tan 2 θ + a 2 d θ = ∫ sec θ d θ = ln ( tan θ + sec θ ) + C tan θ = x a sec θ = sec ( tan − 1 x a ) = ± tan 2 θ + 1 = ± ( x a ) 2 + 1 = ln ( x a + x 2 a 2 + 1 ) + C = ln ( x a + 1 a x 2 + a 2 ) + C = ln ( x + x 2 + a 2 a ) + C {\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}}}
