三角换元法

在积分

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

中，我们可以用以下的代换

${\displaystyle x=a\sin \theta ,\ dx=a\cos \theta \,d\theta }$

${\displaystyle \theta =\arcsin {\frac {x}{a}}}$

这样，积分变为：

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

注意以上的步骤需要${\displaystyle a>0}$和${\displaystyle \cos \theta >0}$；我们可以选择${\displaystyle a}$为${\displaystyle a^{2}}$的算术平方根，然后用反正弦函数把${\displaystyle \theta }$限制为${\displaystyle -{\frac {\pi }{2}}<\theta <{\frac {\pi }{2}}}$。

对于定积分的计算，我们必须知道积分限是怎样变化。例如，当${\displaystyle x}$从0增加到${\displaystyle {\frac {a}{2}}}$时，${\displaystyle \sin \theta }$从0增加到${\displaystyle {\frac {1}{2}}}$，所以${\displaystyle \theta }$从0增加到${\displaystyle {\frac {\pi }{6}}}$。因此，我们有：

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

在积分

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

中，我们可以用以下的代换：

${\displaystyle x=a\tan \theta ,\ dx=a\sec ^{2}\theta \,d\theta }$

${\displaystyle \theta =\arctan {\frac {x}{a}}}$

这样，积分变为：

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

（a > 0）。

以下的积分

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

可以用部分分式的方法来计算，但是，

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

则必须要用换元法：

${\displaystyle x=a\sec \theta ,\ dx=a\sec \theta \tan \theta \,d\theta }$

${\displaystyle \theta =\operatorname {arcsec} {\frac {x}{a}}}$

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

对于含有三角函数的积分，可以用以下的代换：

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

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

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

${\displaystyle {\begin{aligned}\int {\frac {\cos x}{(1+\cos x)^{3}}}\,dx&=\int {\frac {2}{1+u^{2}}}{\frac {\frac {1-u^{2}}{1+u^{2}}}{\left(1+{\frac {1-u^{2}}{1+u^{2}}}\right)^{3}}}\,du\\&={\frac {1}{4}}\int (1-u^{4})\,du\\&={\frac {1}{4}}\left(u-{\frac {1}{5}}u^{5}\right)+C\\&={\frac {(1+3\cos x+\cos ^{2}x)\sin x}{5(1+\cos x)^{3}}}+C\end{aligned}}}$

• 正切半角公式