# 三角換元法

## 含有a2 − x2的積分

${\displaystyle \int {\frac {dx}{\sqrt {a^{2}-x^{2}}}}}$

${\displaystyle x=a\sin(\theta ),\ dx=a\cos(\theta )\,d\theta }$
${\displaystyle \theta =\arcsin \left({\frac {x}{a}}\right)}$

${\displaystyle \int {\frac {dx}{\sqrt {a^{2}-x^{2}}}}=\int {\frac {a\cos(\theta )\,d\theta }{\sqrt {a^{2}-a^{2}\sin ^{2}(\theta )}}}=\int {\frac {a\cos(\theta )\,d\theta }{\sqrt {a^{2}(1-\sin ^{2}(\theta ))}}}}$
${\displaystyle {}=\int {\frac {a\cos(\theta )\,d\theta }{\sqrt {a^{2}\cos ^{2}(\theta )}}}=\int d\theta =\theta +C=\arcsin \left({\frac {x}{a}}\right)+C}$

${\displaystyle \int _{0}^{\frac {a}{2}}{\frac {dx}{\sqrt {a^{2}-x^{2}}}}=\int _{0}^{\frac {\pi }{6}}d\theta ={\frac {\pi }{6}}.}$

## 含有a2 + x2的積分

${\displaystyle \int {\frac {dx}{a^{2}+x^{2}}}}$

${\displaystyle x=a\tan(\theta ),\ dx=a\sec ^{2}(\theta )\,d\theta }$
${\displaystyle \theta =\arctan \left({\frac {x}{a}}\right)}$

{\displaystyle {\begin{aligned}&{}\quad \int {\frac {dx}{a^{2}+x^{2}}}=\int {\frac {a\sec ^{2}(\theta )\,d\theta }{a^{2}+a^{2}\tan ^{2}(\theta )}}=\int {\frac {a\sec ^{2}(\theta )\,d\theta }{a^{2}[1+\tan ^{2}(\theta )]}}\\&{}=\int {\frac {a\sec ^{2}(\theta )\,d\theta }{a^{2}\sec ^{2}(\theta )}}=\int {\frac {d\theta }{a}}={\frac {\theta }{a}}+C={\frac {1}{a}}\arctan \left({\frac {x}{a}}\right)+C\end{aligned}}}

a > 0）。

## 含有x2 − a2的積分

${\displaystyle \int {\frac {dx}{x^{2}-a^{2}}}}$

${\displaystyle \int {\sqrt {x^{2}-a^{2}}}\,dx}$

${\displaystyle x=a\sec(\theta ),\ dx=a\sec(\theta )\tan(\theta )\,d\theta }$
${\displaystyle \theta =\operatorname {arcsec} \left({\frac {x}{a}}\right)}$
{\displaystyle {\begin{aligned}&{}\quad \int {\sqrt {x^{2}-a^{2}}}\,dx=\int {\sqrt {a^{2}\sec ^{2}(\theta )-a^{2}}}\cdot a\sec(\theta )\tan(\theta )\,d\theta \\&{}=\int {\sqrt {a^{2}(\sec ^{2}(\theta )-1)}}\cdot a\sec(\theta )\tan(\theta )\,d\theta =\int {\sqrt {a^{2}\tan ^{2}(\theta )}}\cdot a\sec(\theta )\tan(\theta )\,d\theta \\&{}=\int a^{2}\sec(\theta )\tan ^{2}(\theta )\,d\theta =a^{2}\int \sec(\theta )\ (\sec ^{2}(\theta )-1)\,d\theta \\&{}=a^{2}\int (\sec ^{3}(\theta )-\sec(\theta ))\,d\theta .\end{aligned}}}

## 含有三角函數的積分

${\displaystyle \int f(\sin x,\cos x)\,dx=\int {\frac {1}{\pm {\sqrt {1-u^{2}}}}}f\left(u,\pm {\sqrt {1-u^{2}}}\right)\,du,\qquad \qquad u=\sin x}$
${\displaystyle \int f(\sin x,\cos x)\,dx=\int {\frac {-1}{\pm {\sqrt {1-u^{2}}}}}f\left(\pm {\sqrt {1-u^{2}}},u\right)\,du\qquad \qquad u=\cos x}$
${\displaystyle \int f(\sin x,\cos x)\,dx=\int {\frac {2}{1+u^{2}}}f\left({\frac {2u}{1+u^{2}}},{\frac {1-u^{2}}{1+u^{2}}}\right)\,du\qquad \qquad u=\tan {\frac {x}{2}}}$
${\displaystyle \int {\frac {\cos x}{(1+\cos x)^{3}}}\,dx=\int {\frac {2}{1+u^{2}}}{\frac {\frac {1-u^{2}}{1+u^{2}}}{\left(1+{\frac {1-u^{2}}{1+u^{2}}}\right)^{3}}}\,du=}$
${\displaystyle {\frac {1}{4}}\int (1-u^{4})\,du={\frac {1}{4}}\left(u-{\frac {1}{5}}u^{5}\right)+C={\frac {(1+3\cos x+\cos ^{2}x)\sin x}{5(1+\cos x)^{3}}}+C}$