三角换元法

三角换元法是一种计算积分的方法，是换元积分法的一个特例。 在积分 ∫ d x a 2 − x 2 {\displaystyle \int {\frac {dx}{\sqrt {a^{2}-x^{2}}}}} 中，我们可以用以下的代换 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}}} 这样，积分变为： ∫ 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}}} 注意以上的步骤需要 a > 0 {\displaystyle a>0} 和 cos ⁡ θ > 0 {\displaystyle \cos \theta >0} ；我们可以选择 a {\displaystyle a} 为 a 2 {\displaystyle a^{2}} 的算术平方根，然后用反正弦函数把 θ {\displaystyle \theta } 限制为 − π 2 < θ < π 2 {\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 x {\displaystyle x} 从0增加到 a 2 {\displaystyle {\frac {a}{2}}} 时， sin ⁡ θ {\displaystyle \sin \theta } 从0增加到 1 2 {\displaystyle {\frac {1}{2}}} ，所以 θ {\displaystyle \theta } 从0增加到 π 6 {\displaystyle {\frac {\pi }{6}}} 。因此，我们有： ∫ 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}}.} 在积分 ∫ d x a 2 + x 2 {\displaystyle \int {\frac {dx}{a^{2}+x^{2}}}} 中，我们可以用以下的代换： 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}}} 这样，积分变为： ∫ 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）。 以下的积分 ∫ d x x 2 − a 2 {\displaystyle \int {\frac {dx}{x^{2}-a^{2}}}} 可以用部分分式的方法来计算，但是， ∫ x 2 − a 2 d x {\displaystyle \int {\sqrt {x^{2}-a^{2}}}\,dx} 则必须要用换元法： 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}}} 对于含有三角函数的积分，可以用以下的代换： ∫ 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