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。 此條目沒有列出任何參考或來源。 (2024年6月20日) 在積分 ∫ d x a 2 − x 2 {\displaystyle \int {\frac {dx}{\sqrt {a^{2}-x^{2}}}}} 中，我們可以用以下的代換 x = a sin θ , d x = a cos θ d θ {\displaystyle x=a\sin \theta ,\ dx=a\cos \theta \,d\theta } θ = arcsin x a {\displaystyle \theta =\arcsin {\frac {x}{a}}} 這樣，積分變為： ∫ d x a 2 − x 2 = ∫ a cos θ d θ a 2 − a 2 sin 2 θ = ∫ a cos θ d θ a 2 ( 1 − sin 2 θ ) = ∫ a cos θ d θ a 2 cos 2 θ = ∫ d θ = θ + C = arcsin x a + C {\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}}} 注意以上的步驟需要 a > 0 {\displaystyle a>0} 和 cos θ > 0 {\displaystyle \cos \theta >0} ；我們可以選擇 a {\displaystyle a} 為 a 2 {\displaystyle a^{2}} 的算術平方根，然後用反正弦函數把 θ {\displaystyle \theta } 限制為 − π 2 < θ < π 2 {\displaystyle -{\frac {\pi }{2}}<\theta <{\frac {\pi }{2}}} 。 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 x {\displaystyle x} 從0增加到 a 2 {\displaystyle {\frac {a}{2}}} 時， sin θ {\displaystyle \sin \theta } 從0增加到 1 2 {\displaystyle {\frac {1}{2}}} ，所以 θ {\displaystyle \theta } 從0增加到 π 6 {\displaystyle {\frac {\pi }{6}}} 。因此，我們有： ∫ 0 a 2 d x a 2 − x 2 = ∫ 0 π 6 d θ = π 6 . {\displaystyle \int _{0}^{\frac {a}{2}}{\frac {dx}{\sqrt {a^{2}-x^{2}}}}=\int _{0}^{\frac {\pi }{6}}d\theta ={\frac {\pi }{6}}.} 在積分 ∫ d x a 2 + x 2 {\displaystyle \int {\frac {dx}{a^{2}+x^{2}}}} 中，我們可以用以下的代換： x = a tan θ , d x = a sec 2 θ d θ {\displaystyle x=a\tan \theta ,\ dx=a\sec ^{2}\theta \,d\theta } θ = arctan x a {\displaystyle \theta =\arctan {\frac {x}{a}}} 這樣，積分變為： ∫ d x a 2 + x 2 = ∫ a sec 2 θ d θ a 2 + a 2 tan 2 θ = ∫ a sec 2 θ d θ a 2 [ 1 + tan 2 θ ] = ∫ a sec 2 θ d θ a 2 sec 2 θ = ∫ d θ a = θ a + C = 1 a arctan x a + C {\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）。 以下的積分 ∫ d x x 2 − a 2 {\displaystyle \int {\frac {dx}{x^{2}-a^{2}}}} 可以用部分分式的方法來計算，但是， ∫ x 2 − a 2 d x {\displaystyle \int {\sqrt {x^{2}-a^{2}}}\,dx} 則必須要用代換法： x = a sec θ , d x = a sec θ tan θ d θ {\displaystyle x=a\sec \theta ,\ dx=a\sec \theta \tan \theta \,d\theta } θ = arcsec x a {\displaystyle \theta =\operatorname {arcsec} {\frac {x}{a}}} ∫ x 2 − a 2 d x = ∫ a 2 sec 2 θ − a 2 ⋅ a sec θ tan θ d θ = ∫ a 2 ( sec 2 θ − 1 ) ⋅ a sec θ tan θ d θ = ∫ a 2 tan 2 θ ⋅ a sec θ tan θ d θ = ∫ a 2 sec θ tan 2 θ d θ = a 2 ∫ sec θ ( sec 2 θ − 1 ) d θ = a 2 ∫ ( sec 3 θ − sec θ ) d θ . {\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}}} 對於含有三角函數的積分，可以用以下的代換： ∫ f ( sin x , cos x ) d x = ∫ 1 ± 1 − u 2 f ( u , ± 1 − u 2 ) d u , u = sin x {\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} ∫ f ( sin x , cos x ) d x = ∫ − 1 ± 1 − u 2 f ( ± 1 − u 2 , u ) d u u = cos 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} ∫ f ( sin x , cos x ) d x = ∫ 2 1 + u 2 f ( 2 u 1 + u 2 , 1 − u 2 1 + u 2 ) d u u = tan x 2 {\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}}} ∫ cos x ( 1 + cos x ) 3 d x = ∫ 2 1 + u 2 1 − u 2 1 + u 2 ( 1 + 1 − u 2 1 + u 2 ) 3 d u = 1 4 ∫ ( 1 − u 4 ) d u = 1 4 ( u − 1 5 u 5 ) + C = ( 1 + 3 cos x + cos 2 x ) sin x 5 ( 1 + cos x ) 3 + C {\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 Wikiwand in your browser!Seamless Wikipedia browsing. 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![{\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}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/22887b99fc50663f4f90ae0b609bd4a2e3d2109e)
![{\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}}}](http://wikimedia.org/api/rest_v1/media/math/render/svg/22887b99fc50663f4f90ae0b609bd4a2e3d2109e)
