三角代換法是一種計算積分的方法,是代換積分法的一個特例。
含有a2-x2的積分[編輯]
在積分
![{\displaystyle \int {\frac {dx}{\sqrt {a^{2}-x^{2}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b554f29e05d3409df9950063ac8a7c0a0b218e41)
中,我們可以用以下的代換
![{\displaystyle x=a\sin \theta ,\ dx=a\cos \theta \,d\theta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/667f1df15c03cde01fa75c01b79ebfbb17df0fdd)
![{\displaystyle \theta =\arcsin {\frac {x}{a}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3da775284ebc3dfcbdd6904074891a622df940b3)
這樣,積分變為:
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d75dbc74a15f7fda7fe14b203751a08252c633a)
注意以上的步驟需要
和
;我們可以選擇
為
的算術平方根,然後用反正弦函數把
限制為
。
對於定積分的計算,我們必須知道積分限是怎樣變化。例如,當
從0增加到
時,
從0增加到
,所以
從0增加到
。因此,我們有:
![{\displaystyle \int _{0}^{\frac {a}{2}}{\frac {dx}{\sqrt {a^{2}-x^{2}}}}=\int _{0}^{\frac {\pi }{6}}d\theta ={\frac {\pi }{6}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a3ef74b44c38129edf5862ce3c188301498a083)
含有a2+x2的積分[編輯]
在積分
![{\displaystyle \int {\frac {dx}{a^{2}+x^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1c09e16acbfd7c4c1079c381944b55247f8feada)
中,我們可以用以下的代換:
![{\displaystyle x=a\tan \theta ,\ dx=a\sec ^{2}\theta \,d\theta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd44cadce9da22530d659712b58a1ae74fac732b)
![{\displaystyle \theta =\arctan {\frac {x}{a}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/893cb3b3d548466285e691f2c40409c6ca0e9be8)
這樣,積分變為:
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22887b99fc50663f4f90ae0b609bd4a2e3d2109e)
(a > 0)。
含有x2 − a2的積分[編輯]
以下的積分
![{\displaystyle \int {\frac {dx}{x^{2}-a^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9b30f75798064140e18d96a88c285fd35808652c)
可以用部分分式的方法來計算,但是,
![{\displaystyle \int {\sqrt {x^{2}-a^{2}}}\,dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b0adc36a27b13417d665fc5e270f00f76aef98dd)
則必須要用代換法:
![{\displaystyle x=a\sec \theta ,\ dx=a\sec \theta \tan \theta \,d\theta }](https://wikimedia.org/api/rest_v1/media/math/render/svg/571a1da8a9b91b34fa0d7bc5f73371290e101609)
![{\displaystyle \theta =\operatorname {arcsec} {\frac {x}{a}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/34233ecf0599db7b3eb44ca5e8fc5308b150ff8a)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dfc08229a2b4bdcc5506c325804b84e1f3d81653)
含有三角函數的積分[編輯]
對於含有三角函數的積分,可以用以下的代換:
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d1a7fde32a93f1d946121646e32d8239cf0f9f0d)
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0f95fa7200a0117b854fe4caed46eb0a554562b6)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a5557221e5f8a18e88d4de46c8b36f5ee3276da7)
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/846ed9eb22e60e90fcf2e8129432d2554240b1de)